# Concept Of Physics Simple Harmonics Motion

#### H C Verma

1.   The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes $2$ cm, $1$ m/s and $10$ m/s $2$ at a certain instant. Find the amplitude and the time period of the motion.

Given that, at a particular instance,

$x$ =$2cm$ = $0.02m$

$v$ = $1$$m/sec A = 10$$msec^{-2}$

We Know that $a$ = $\omega^{2}$$x \Rightarrow \omega = \sqrt\frac{a}{x} = \sqrt\frac{10}{0.02} = \sqrt500 = 10\sqrt5 T = \frac{2\pi}{\omega} = \frac{2}{10\sqrt5} = \frac{2x3.14}{10x2.236} = 0.28seconds. Again, amplitude r is given by v = \omega (\sqrt r^{2}-x^{2}) \omega \Rightarrow v^{2} = \omega^{2}(r^{2}-x^{2}) 1 = 500 (r^{2}-0.0004) \Rightarrow r = 0.0489 \approx0.049m \therefore r = 4.9 cm 2. A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6s. At t = 0 it is at position x= 5 cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t = 4 s. Given r = 10cm At t = 0,x =5cm so, w = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3} sec^{-1} At t = 0,x = 5cm so, 5 = 10 sin (wx0x\phi) = 10 sin \phi \\ [y = r sin wt] sin \phi = 1/2 \Rightarrow \phi = \frac{\pi}{6} therefore Equation of displacement x = (10cm) sin \big(\frac{\pi}{3}\big) (ii) At t = 4 seconds x = 10 sin \big[\frac{\pi}{3} x 4 + \frac{\pi}{6}\big] = 10 sin \big[\frac{8\pi + \pi}{6}\big] = 10 sin \big(\frac{3\pi}{2}\big) = 10 sin \big(\pi + \frac{\pi}{2}\big) = -10 sin \big(\frac{\pi}{2}\big) = -10 Acceleration a = - w^{2} x = - \big(\frac{\pi^2}{9}\big) x (-10) = 10.9 \approx 0.11 cm/sec 3. A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from -the mean position are the kinetic and potential energies equal ? r = 10cm Bacause, K.E. = P.E. so (1/2) m \omega^{2} (r^{2}- y^{2}) = (1/2) m \omega^{2}$$y^{2}$

so (1/2) m $\omega^{2}$ ($r^{2}$-$y^{2}$) = (1/2) m $\omega^{2}$$y^{2} r^{2} - y^{2} = y^{2} \Rightarrow 2y^{2} = r^{2} \Rightarrow y = \frac{r}{\sqrt2} = \frac{10}{\sqrt2} = 5\sqrt2 cm form the mean position 4. The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm/s and 50 cm/s^{2}. Find the position(s) of the particle when the speed is 8 cm/s. Answer 4 None V_{max} = 10 cm/sec \Rightarrow r \omega = 10 \Rightarrow \omega^{2} = \frac{100}{r^{2}} ......(1) A_{max} = \omega^{2}$$r$ = $50$ $cm/sec$

$\Rightarrow$ $\omega^{2}$ = $\frac{50}{y}$ = $\frac{50}{r}$ ......(2)

$\therefore$ $\frac{100}{r^2}$ = $\frac{100}{r}$ $\Rightarrow$ $r$ = $2cm$

$\therefore$ $\omega$ = $\sqrt\frac{100}{r^2}$ = $5$ $sec^{2}$

Again to find out the position where the speed is $8m/sec$.

$V^{2}$ = $\omega^{2}$ ($r^{2}$-$y^{2}$)

$\Rightarrow$ $64$ = $25$ (4 - $y^{2}$)

$\Rightarrow$ (4 - $y^{2}$) = $\frac{64}{25}$ $\Rightarrow$ $y^{2}$ = $1.44$ $\Rightarrow$ $y$ = $\pm$ $1.2$ $cm$ from mean position.

5.   A particle having mass 10 g oscillates according to the equation $x$ = ($2.0 cm$) sin[($100$ s$^{-1}$ )t + $\pi$/6]. Find (a) the amplitude, the time period and the spring constant (b) the position, the velocity and the acceleration at $t$ = $0$.

$x$ = $(2.0cm)$ sin [(100$s^{-1}$) t + ($\pi$/6)

$m$ =$10g$

$a)$ Amplitude = $2cm$

$\omega$ = 100$s^{-1}$

$\therefore$ $T$ = $\frac{2\pi}{100}$ = $\frac{\pi}{50}$$sec = 0.063 sec We know that T = 2\pi \sqrt\frac{m}{k} \Rightarrow T^{2} = 4\pi^{2} x \frac{m}{k} \Rightarrow k = \frac{4\pi^2}{T^2}m = 10^5 dyne/cm = 100 N/M. \\ [because \omega = \frac{2\pi}{T} = 100sec^-1] b) At t = 0 x = 2 cm sin \big(\frac{\pi}{6}\big) = 2 x (1/2) = 1 cm.from the mean position. we know that x = A sin (\omega t + \phi) v = A cos (0 + \frac{\pi}{6}) = 200 x \frac{\sqrt3}{2} = 100 \sqrt3 sec^{-1} = 1.73m/s c) a = -\omega^2 x =100^2 x 1 = 100 m/s^2 6. The equation of motion of a particle started at t = 0 is given by x = 5 sin (20 t + n/3) w here x is in centim etre and t in second. When does the particle\\ (a) first come to rest \\ (b) first have zero acceleration \\ (c) first have maximum speed ? Answer 6 None x = 5 sin (20t + \pi/3) a) Max. displacement from the mean position = Amplitude of the particle. At the extream position, the velocity becomes '0'. \therefore x = 5 = Amplitude. \therefore 5 = 5 sin (20t + \pi/3) sin (20t + \pi/3) = 1 = sin (\pi/2) \Rightarrow 20t + \pi/3 = \pi/2 \Rightarrow t = \pi/120 sec it first comes to rest. b) a = \omega^2 x = \omega^2 [5 sin (20t + \pi/3)] For a = 0,5 sin (20t + \pi/3) = 0 \Rightarrow sin (20t + \pi/3) = sin (\pi) \Rightarrow 20t = \pi - \pi/3 = 2\pi/3 \Rightarrow t = \pi/30 sec. c) v = A \omega cos (\omega t + \pi/3) when, v is maximum i.e cos (20t + \pi/3) = -1 = cos \pi \Rightarrow 20t = \pi - \pi/3 = 2\pi/3 \Rightarrow t = \pi/30 sec. 7. Consider a particle moving in simple harmonic motion according to the equation x = 2.0 cos (50 \pi t + tan^{-1} - 0.75) where x is in centimetre and t in second. The motion is started at t = 0. \\(a) When does the particle come to rest for the first time ? \\(b) When does the acceleration have its maximum magnitude for the first time ? \\(c) When does the particle come to rest for the second time ? a) x = 2.0 cos (50$$\pi$t = $tan^{-1}$ $0.75$)

$a)$ $x$ = $2.0$ $cos$ ($50$$\pit = tan^{-1} 0.75) = 2.0 cos (50$$\pi$t + $0.643$)

$v$ = $\frac{dv}{dt}$ = - $100$ $sin$ (50 $\pi$ t + $tan^{-1}$ - 0.75) = $2.0$ $cos$ (50$\pi$t + $0.643$)

$\Rightarrow$ $sin$ (50$\pi$t + $0.643$) = $0$

As the particle comes to rest for the $1^{st}$ time

$\Rightarrow$ 50$\pi$t + $0.643$ = $\pi$

$t$ = $1.6$ x $10^{-2}$ $sec$

$b)$ Acceleration $a$ = $\frac{dv}{dt}$ = - $100$$\pi x 50 \pi cos (50 \pi t + 0.643) For Maximum acceleration cos (50 \pi t + 0.643) = -1 cos \pi (max)(so a is max) \Rightarrow t = 1.6 x 10^{-2} sec c) When the particle comes to rest for second time, 50 \pi t + 0.643 = 2$$\pi$

$\Rightarrow$ $t$ = $3.4$ x $10^{-2}$ $s$.

8.   Consider a simple harmonic motion of time period T. Calculate the time taken for the displacement to change value from half the amplitude to the amplitude.

$y_1$ = $\frac{r}{2}$, $y_2$ = $r$ (for the two given position)

Now, $y_1$ = $r$ $sin$ $\omega$$t_1 \Rightarrow \frac{r}{2} = r sin \omega$$t_{1}$ $\Rightarrow$ $sin$ $\omega$$t_1 = \frac{1}{2} \Rightarrow \omega$$t_{1}$ = $\frac{\pi}{2}$ $\Rightarrow$ $\frac{2\pi}{t}$ x $t_1$ = $\frac{\pi}{6}$ $\Rightarrow$ $t_1$ - $\frac{t}{12}$

Again, $y_2$ = $r$ $sin$ $\omega$$t_2 \Rightarrow r = r sin \omega$$t_{2}$ $\Rightarrow$ $sin$ $\omega$$t_2 = 1 \Rightarrow \omega$$t_{2}$ = $\frac{\pi}{2}$ $\Rightarrow$ $\big(\frac{2\pi}{t}\big)$ $t_2$ = $\frac{\pi}{2}$ $\Rightarrow$ $t_2$ - $\frac{t}{4}$

so, $t_2$ - $t_1$ = $\frac{t}{4}$ -$\frac{t}{12}$ = $\frac{t}{6}$

9.   The pendulum of a clock is replaced by a spring-mass system with the spring having spring constant $0$'$1$ N/m. What mass should be attached to the spring ?

$k$ = $0.1$ N/M T = $2\pi$ $\sqrt\frac{m}{k}$ = $2$ $sec$ [Time period of pendulum of a clock = $2$ $sec$]

So, $4$$\pi^2 + \big(\frac{m}{k}\big) = 4 \therefore m = \frac{k}{\pi^2} = \frac{0.1}{10} = 0.01kg \approx 10 gm 10. A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals the length of an equivalent simple pendulum i.e., a pendulum having frequency same as that of the block Time period of simple pendulum = 2\pi \sqrt\frac{1}{g} Time period of spring is 2\pi \sqrt\frac{m}{k} T_p = T_s [Frequency is same] \Rightarrow \sqrt\frac{1}{g} = \sqrt\frac{m}{k} \Rightarrow \frac{1}{g} = \frac{m}{k} \Rightarrow 1 = \frac{mg}{k} = \frac{F}{K} = x. (Because , restoring force = weight = F =msg) \Rightarrow 1 = x (proved) 11. A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 m and time period 0'314 s. Find the maximum force exerted by .the spring on the block. x = r = 0.1 m T = 0.314 sec m = 0.5 kg. Total force exerted on the block = weight of the block + spring force. T = 2$$\pi$$\sqrt\frac{m}{k} \Rightarrow 0.314 = 2\pi$$\sqrt\frac{0.5}{k}$ $\Rightarrow$ $k$ = $200$ N/m

$\therefore$ Force exerted by the spring on the block is

$f$ = $Kx$ = $201.1$ x $0.1$ = $20$N

$\therefore$ Maximum force = $f$ + weight = $20$ = $5$ = $25$N

12.   A body of mass $2$ kg suspended through a vertical spring executes simple harmonic motion of period $4$ s. If the oscillations are stopped and the body hangs in equilibrium, find the potential energy stored in the spring.

$m$ = $2$$kg. T =4$$sec$.

$T$ = $2$$\pi$$\sqrt\frac{m}{k}$ $\Rightarrow$ $4$ = $2$$\pi$$\sqrt\frac{2}{k}$ $\Rightarrow$ $2$ = $\pi$$\sqrt\frac{2}{k} \Rightarrowb4 = 2$$\pi$$\sqrt\frac{2}{k} \Rightarrow 2 = \pi$$\sqrt\frac{2}{k}$

$\Rightarrow$ $4$ = $\pi^2$$\big(\frac{2}{k}\big) \Rightarrow k = \frac{2\pi^2}{4} \Rightarrow k = \frac{\pi^2}{2} = 5 N/M But, we know that f = mg = kx \Rightarrow x = \frac{mg}{k} = \frac{2\times10}{5} = 4 \therefore Potential Energy = (1/2) k x^2 = (1/2) x 5x16 = 5 x 8 = 40 J 13. A spring stores 5 J of energy when stretched by 25 cm. It is kept vertical with the lower end fixed. A block fastened to its other end is made to undergo small oscillations. If the block makes 5 oscillations each second, what is the mass of the block ? x = 25 cm = 0.25$$m$

$E$ = $5j$

$f$ = $5$

So, $T$ = $1/5$$sec. Now P.E. = (1/2) Kx^2 \Rightarrow(1/2)kx^2 = 5 \Rightarrow (1/2) k (0.25)^{2} = 5 =\Rightarrow K = 160 N/m. Again, T = 2$$\pi$$\sqrt\frac{m}{k} \Rightarrow \frac{1}{5} = 2$$\pi$$\sqrt\frac{m}{160} \Rightarrow m = 0.16$$kg$

14.   A small block of mass m is kept on a bigger block of mass M which is attached to a vertical spring of spring constant k as shown in the figure. The system oscillates vertically. $\\$(a) Find the resultant force on the smaller block when it is displaced through a distance x above its equilibrium position. $\\$ (b) Find the normal force on the smaller block at this position. When is this force smallest in magnitude ? $\\$(c) What can be the maximum amplitude with which the two blocks may oscillate together ?

$a)$ From the free body diagram,

$\therefore$ $R$ + $m\omega^2x$ - $mg$ =$0$ ......(1) $\\$ Resultant force $m\omega^2x$ = $mg$ - $R$

$\Rightarrow$ $m\omega^2x$ = $m$$\big(\frac{k}{M+m}\big) \Rightarrow x = \frac{mkx}{M+m} [ \omega = \sqrt{K/(M+m)} for spring mass system ] b) R = mg - m\omega^2x = mg - m$$\frac{k}{M+m}$$x = mg - \frac{mkx}{M+m} For R to be smallest m\omega^2x should be max i.e. x is maximum.\\ The particle shoild be at the high point. c) We have R = mg - m\omega^2x The two blocks may oscillates together is such a way that R is grater than 0.At limiting condition,\\ R = 0, mg = m\omega^2x X = \frac{mg}{m\omega^2} = \frac{mg(M+m)}{mk} So, the maximum amplitude is = \frac{g(M+m)}{k} 15. The block of mass in, shown in figure (12-E2) is fastened to the spring and the block of mass m_2 is placed against it. \\ (a) Find the compression of the spring in the equilibrium position. \\ (b) The blocks are pushed a further distance (2/h) (m1+ m2)g sin\theta against the spring and released. Find the position where the two blocks separate.\\ (c) What is the common speed of blocks at the time of separation ? a) At the equilibrium condition, kx = (m_1 + m_2)g sin \theta (Given) \Rightarrow x = \frac{(m_1 + m_2)g sin \theta}{k} b) x_1 = \frac{2}{k}$$(m_1 + m_2)$ $g$ $sin\theta$ (Given)

when the system is released, it will start to make SHM

where $\omega$ = $\sqrt\frac{k}{m_1+m_2}$

When the block lose contact, $P$ = $0$

So $m_2$ $g$ $sin$ $\theta$ = $m_2$ $x_2$ $\omega^2$ = $m_2x_2$ $\big(\frac{k}{m_1+m_2}\big)$

$\Rightarrow$ $x_2$ = x $k$$(m_1 + m_2)g sin \theta So the block will lose contact with each other when the springs attain its natural length. c) Let the common speed attained by both the blocks be v. 1/2 (m_1+m_2)$$v^2$ - $0$ = $1/2$ $k(x_1 + x_2)^{2}$ - $(m_1+m_2)$ $g$ $sin$ $\theta$ $(x + x_1)$ [$x + x_1$ = total compression]

$\Rightarrow$ $1/2$ $(m_1+m_2)$$v^2 \Rightarrow (1/2) (m_1+m_2) v^2 = [(1/2) k (3/k) ((m_1 + m_2)g sin \theta - (m_1 + m_2)g sin \theta] (x + x_1) \Rightarrow (1/2) (m_1+m_2) v^2 = [(1/2) (m_1 + m_2)g sin \theta x (3/k) (m_1 + m_2)g sin \theta \Rightarrow v = \sqrt\frac{3}{k(m_1+m_2)} g sin$$\theta$

$1/2$ $(m_1+m_2)$$v^2 - 0 = 1/2 k(x_1 + x_2)^{2} - (m_1+m_2) g sin \theta (x + x_1) [x + x_1 = total compression] 16. A particle of mass in is attatched to three springs A, B and C of equal force constants k as shown in figure (12-E6). If the particle is pushed slightly against the spring C and released, find the time period of oscillation. Give, k = 100 N/m, \\ m = 1kg and F = 10 N a) In the euilibrium position, \\ compression \delta = F/k = 10/100 = 0.1m = 10 cm b) The below imparts a speed of 2m/s to block towards left. \therefore P.E. + K.E. = 1/2 k\delta^2 + 1/2 Mv^2 = (1/2) x 100 x (0.1)^2 + (1/2) x 1 x 4 = 0.5 + 2 = 2.5 J c) Time period = 2\pi$$\sqrt{\frac{M}{k}}$ = $2\pi$$\sqrt{\frac{1}{100}} = \frac{\pi}{5} sec d) Let the amplitude be'x' which means the distance between the mean position and the extream position. So, in the extream position, compression of the spring is (x+\delta). Since, in SHM,the total enegy remaining constant. So, in the extreme position, compression of the spring is (x+\delta). So, 50(x+0.1)^{2} = 2.5 +10x \therefore 50x^{2} + 0.5 + 10x = 2.5 + 10x \therefore 50x^{2} = 2 \Rightarrow x^{2} = \frac{2}{50} = \frac{4}{100} \Rightarrow x = \frac{2}{10}m = 20$$cm$.

$e)$ Potential Energy at the left extreme is given by.

$P.E$ = $(1/2)$$k (x+\delta)^{2} = (1/2) x 100(0.1+0.2)^{2} = 50 x0.09 = 4.5 J f) Potential Energy at the right extream is given by, P.E = (1/2)$$k$ $(x+\delta)^{2}$ - $F(2x)$ $\\$ [$2x$ = distance between two extremes]

So, in the extreme position, compression of the spring is $(x+\delta)$.

Since, in SHM,the total energy remaining constant.

$(1/2)$$k$$(x+\delta)^{2}$ = $(1/2)$$k\delta^{2}+(1/2)$$mv^{2}$+$Fx$ = $2.5$ +$10$$x \\ [because (1/2)$$k\delta^{2}$+$(1/2)mv^{2}$ = $2.5$]

So, $50(x+0.1)^{2}$ = $2.5$ +$10x$

$\therefore$ $50x^{2}$ + $0.5$ + $10x$ = $2.5$ + $10x$ $\\$ $\therefore$ $50x^{2}$ = $2$ $\Rightarrow$ $x^{2}$ = $\frac{2}{50}$ = $\frac{4}{100}$ $\Rightarrow$ $x$ = $\frac{2}{10}$m = $20$$cm. e) Potential Energy at the left extreme is given by. P.E = (1/2)$$k$ $(x+\delta)^{2}$ = $(1/2)$ x $100(0.1+0.2)^{2}$ = $50$ x$0.09$ = $4.5$ $J$

$f)$ Potential Energy at the right extream is given by,

$P.E$ = $(1/2)$$k (x+\delta)^{2} - F(2x) \\ [2x = distance between two extremes] = 4.5 - 10(0.4) = 0.5$$J$

The different value in $(b)$ $(e)$ and $(f)$ do not violate law of conservation of energy as the work is done by $\\$ the external force $10$N.

17.   Find the time period of the oscillation of mass m in figures $(12-$E4 a, b, c$)$ What is the equivalent spring constant of the pair of springs in each case ?

$a)$ Equivalent spring constant = $k$ = $k_1$ + $k_2$ (parallel)

$T$ = $2\pi$ $\frac{M}{K}$ = $2\pi$ $\sqrt{\frac{M}{K_1+K_2}}$

$b)$ Let us, displace the block $m$ towards left through displacement '$x$'

Resultant force $F$ = $F_1+F_2$ = ($K_1+K_2$)$x$

Time period $T$ = $2\pi$ $\sqrt{\frac{displacement}{Acceleration}}$ = $2\pi$ $\sqrt{\frac{x}{\frac{m(K_1+K_2)}{m}}}$ =$2\pi$ $\sqrt{\frac{M}{K_1+K_2}}$

The equivalent spring constant $k$ = $k_1$ + $k_2$

$c)$ In series conn equivalent spring constant be $k$. So, $\frac{1}{k}$ = $\frac{1}{k_1}$ + $\frac{1}{k_2}$ = $\frac{k_2+k_1}{k_1k_2}$ $\Rightarrow$ $k$ = $\frac{k_1k_2}{k_2+k_1}$

$T$ = $2\pi$ $\frac{M}{k}$ = $2\pi$ $\frac{m(k_1+k_2)}{k_1k_2}$

18.   The spring shown in figure $(12-E5)$ is unstretched when a man starts pulling on the cord. The mass of the block is $M$. If the man exerts a constant force $F$, find $\\$ $(a)$ the amplitude and the time period of the motion of the block, $\\$(b) the energy stored in the spring when the block passes through the equilibrium position and $\\$(c) the kinetic energy of the block at this position.

$a)$ We have $F$ = $kx$ $\Rightarrow$ $x$ = $\frac{F}{k}$

Acceleration = $\frac{F}{m}$

Time period $T$ = $2\pi$ $\sqrt{\frac{displacement}{Acceleration}}$ = $2\pi$ $\sqrt{\frac{F/k}{F/m}}$ =$2\pi$ $\sqrt{\frac{m}{k}}$

Amplitude = max displacement = $F/k$

$b)$ The energy stored in the spring when the block passes through the equilibrium position

$(1/2)kx^2$ = $(1/2)k(F/k)^{2}$ = $(1/2)k(F^2/k^2)$ = $(1/2)$$(F^2/k) c) At the mean position, P.E. is 0.K.E. is (1/2)$$kx^{2}$ = $(1/2)$$(F^2/x) 19. A particle of mass in is attatched to three springs A, B and C of equal force constants k as shown in figure (12-E6). If the particle is pushed slightly against the spring C and released, find the time period of oscillation. Suppose the particle is pushed slightly against the spring 'C' through displacement 'x'. Total resultant force on the particle is kx due to spring C and \frac{kx}{\sqrt2} due to spring A B. \therefore Total Resultant force = kx + \sqrt{(\frac{kx}{\sqrt{2}}) + (\frac{kx}{\sqrt{2}})} = kx + kx = 2kx. Acceleration = \frac{2kx}{m} Tme period T = 2\pi \frac{displacement}{Acceleration} = 2\pi \sqrt{\frac{x}{\frac{2kx}{m}}} = 2\pi$$\sqrt{\frac{m}{2k}}$

$[$Cause $:-$ when the body pushed again '$C$' the spring $C$ , tries to pull the block towards $XL$.At that moment the spring $A$ and $B$ tries to pull the block with force $\frac{kx}{\sqrt2}$ and

$\frac{kx}{\sqrt2}$ respectively towards $xy$ and $xz$ respectively. So the total force on the block is due to the spring force '$C'$ as well as the component of two spring force $A$ and $B$$] 20. Repeat the previous exercise if the angle between each pair of springs is 120° initially. Options 1 F = kx + \frac{kx}{2} = \frac{3kx}{2} Answer 20 None In this case , if the particle 'm' is pushed against 'C' a by distance 'x'.\\ Total resultant force acting on man 'm' is given by, F = kx + \frac{kx}{2} = \frac{3kx}{2} [ Because net force A & B = \sqrt{(\frac{kx}{2})^2+(\frac{kx}{2})^2 + 2(\frac{kx}{2})(\frac{kx}{2})cos120^0} = \frac{kx}{2} \therefore a = \frac{F}{m} = \frac{3km}{2m} \Rightarrow \frac{a}{x} = \frac{3k}{2m} = \omega^2 \sqrt{\frac{3k}{2m}} \therefore Time period T \frac{2\pi}{\omega} = 2\pi$$\sqrt{\frac{2m}{3k}}$

21.   The springs shown in the figure $(12-E7)$ are all unstretched in the beginning when a man starts pulling the block. The man exerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block.

$k_2$ and $k_3$ are in series.

Let equivalent spring constant be $k_4$

$\therefore$ $\frac{1}{k_4}$ = $\frac{1}{k_2}$ + $\frac{1}{k_3}$ = $\frac{k_2+k_3}{k_2k_3}$ $\Rightarrow$ $k_4$ = $\frac{k_2k_3}{k_2+k_3}$

Now $k_4$ and $k_1$ are in parallel.

So equivalent spring constant $k$ = $k_3$+$k_4$ = $\frac{k_2k_3}{k_2+k_3}$ +$k_1$ = $\frac{k_2k_3+k_1k_2+k_1k_3}{k_2+k_3}$

$\therefore$ $T$ = $2\pi$ = $\sqrt{\frac{M}{k}}$ = $2\pi$ $\sqrt{\frac{M(k_2+k_3)}{k_2k_3+k_1k_2+k_1k_3}}$

$b)$ frequency = $\frac{1}{T}$ = $\frac{1}{2\pi}$ $\sqrt{\frac{k_2k_3+k_1k_2+k_1k_3}{M(k_2+k_3)}}$

$c)$ Amplitude $x$ = $\frac{F}{k}$ = $\frac{F(k_2+k_3)}{k_1k_2+k_2k_3+k_1k_3}$

22.   Find the elastic potential energy stored in each spring shown in figure $(12-E8)$, when the block is in equilibrium. Also find the time period of vertical oscillation of the block.

$k_1$$k_2$$k_3$ are in series.

$\frac{1}{k}$ = $\frac{1}{k_1}$ + $\frac{1}{k_2}$ + $\frac{1}{k_3}$ $\\$ $\Rightarrow$ $k$ = $\frac{k_1k_2k_3}{k_1k_2+k_2k_3+k_1k_3}$

Time period $T$ = $2\pi$ $\sqrt{\frac{m}{k}}$ $2\pi$$\sqrt{\frac{k_1k_2k_3}{k_1k_2+k_2k_3+k_1k_3}} = 2\pi$$\sqrt{m(\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3})}$

Now, Force = weight = mg.

$\therefore$ At $k_1$ spring, $x_1$ = $\frac{mg}{k_1}$

Similarly $x_2$ = $\frac{mg}{k_2}$ and $x_3$ = $\frac{mg}{k_3}$

$\therefore$ PE_1 = $(1/2)$$k_1x_1^{2} = \frac{1}{2}k_1$$(\frac{Mg}{k_1})^{2}$ = $\frac{1}{2}k_1$$\frac{m^2g^2}{k_1^{2}} = \frac{m^2g^2}{2k_1} Similarly PE_2 = \frac{m^2g^2}{2k_2} and PE_3 = \frac{m^2g^2}{2k_3} 23. The string, the spring and the pulley shown in figure (12-E9) are light. Find the time period of the mass m. When only 'm' is hanging, let the extension in the spring be 'l' So T_1 = kl = mg. When a force F is applied, let the further extension be'x' \therefore T_2 = k(x+l) \therefore Driving force = T_2 - T_1 = k(x+l) - kl = kx \therefore Acceleration = \frac{kl}{m} T = 2\pi$$\sqrt{\frac{displacement}{Acceleration}}$ = $2\pi$$\sqrt{\frac{x}{\frac{Kx}{m}}} = 2\pi$$\frac{m}{k}$

24.   Solve the previous problem if the pulley has a moment of inertia $I$ about its axis and the string does not slip over it

Let us solve the problem by 'energy method'.

initial extension of the spring in the mean position.

$\delta$ = $\frac{mg}{k}$

During osillation, at any position '$x$' below the equilibrium position, let the velocity of '$m$' be $v$ and $\\$ angular velocity of the pulley be '$\omega$'. if $r$ is the radius of the pulley, then $v$ = $r\omega$. $\\$ At any instant, Total Energy = constant (for SHM)

$\therefore$ $(1/2)mv^2$ + $(1/2)I\omega^2$ + $(1/2)k[(x+\delta)^{2} - \delta^2]$ - $mgx$ = Constant

$\Rightarrow$ $\therefore$ $(1/2)mv^2$ + $(1/2)I\omega^2$ + $(1/2) kx^{2} - kx\delta$ - $mgx$ = Constant

$\Rightarrow$ $\therefore$ $(1/2)mv^2$ + $(1/2)I(v^2/r^2)$ + $(1/2) kx^{2}$ = Constant $\\$ $(\delta = mg/k$)

Taking derived of both sides either respect to '$t$'.

$mv$$\frac{dv}{dt} + \frac{I}{r^{2}}$$V$$\frac{dv}{dt}+kx$$\frac{dv}{dt}$ = $0$

$\Rightarrow$ $a$$\big(m+\frac{I}{r^2}\big) = kx \\ (\therefore x = \frac{dx}{dt} and a = \frac{dx}{dt}) \Rightarrow \frac{a}{x} = \frac{k}{m+\frac{I}{r^2}} = \omega^2 \Rightarrow = T = 2\pi$$\sqrt{\frac{m+\frac{I}{r^2}}{k}}$

25.   Consider the situation shown in figure $(12-E10)$. Show that if the blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. Calculate the time period

The center of mass of the system should not change during the motion. So,if the block '$m$' on the left $\\$ moves towards right a distance '$x$',the block on the right moves left a distance '$x$'.So,total$\\$ compression of the spring is $2x$.

By energy method, $\frac{1}{2}$$k$$(2x)^{2}$+$\frac{1}{2}$$mv^2 + \frac{1}{2}mv^{2} + \frac{1}{2}mv^{2} = C \Rightarrow mv^{2} + 2kx^{2} = C. Taking derivation of both sides with respect to 't'. m x 2v \frac{dv}{dt} + 2k x 2x\frac{dx}{dt} = 0 \therefore ma + 2kx = 0 \\ [because v = dx/dt and a = dv/dt] \Rightarrow \frac{a}{x} = -$$\frac{2k}{m}$ = $\omega^2$ $\omega$ = $\sqrt{\frac{2k}{m}}$

$\Rightarrow$ Time period $T$ = $2\pi$$\sqrt{\frac{m}{2k}} 26. A rectangular plate of sides a and b is suspended from a ceiling by two parallel strings of length L each (figure 12-E11). The separation between the strings is d. The plate is displaced slightly in its plane keeping the strings tight. Show that it will execute simple harmonic motion. Find the time period. Here we have to consider oscillation of center of mass Driving force F = mg sin \theta. Acceleration = a = \frac{F}{m} = g sin \theta For small angle \theta, sin \theta = \theta. \therefore a = g \theta = g(\frac{x}{L}) \\ [where g and L are constant] \therefore a = x_1 So the motion is simple Harmonic Time period T = 2\pi\sqrt{\frac{Displacement}{Acceleration}} = 2\pi$$\sqrt{\frac{x}{(\frac{gx}{L})}}$ = $2\pi$ $\sqrt{\frac{L}{g}}$

27.   A $1$ $kg$ block is executing simple harmonic motion of amplitude $0$'$1$ $m$ on a smooth horizontal surface under the restoring force of a spring of spring constant $100$ $N/m$. A block of mass $3$ $kg$ is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.

Amplitude = $0.1$ $m$

Total mass = $3$ + $1$ = $4kg$ (when both the blocks are moving together)

$\therefore$ $T$ = $2\pi$$\sqrt{\frac{M}{k}} = 2\pi$$\sqrt{\frac{4}{100}}$ = $\frac{2\pi}{5}$ $sec$.

$\therefore$ Frequency = $\frac{5}{2\pi}$ $Hz$.

Again at the mean position, let $1kg$ block has velocity $v$.

KE. = $(1/2)mv^{2}$ = $(1/2)mx^{2}$ $\\$ Where $x$$\rightarrow Amplitude = 0.1 m. \therefore (1/2) x (1 x v^2) = (1/2) x 100 (0.1)^{2} \Rightarrow v = 1$$m/sec$ $..........(1)$

After the $3kg$ block is gentely placed on the $1kg$, then let, $1kg$ + $3kg$ = $4kg$ block and the spring be one $\\$ system.For this mass spring system, there is do external force. (when oscillation takes place). the $\\$ The momentum should be conserved. Let, $4kg$ block has velocity $v^|$.

$\therefore$ Initial momentum = Final momentum

$\therefore$ $1$ x $v$ = $4$ x $v^{|}$ $\Rightarrow$ $v^{|}$ = $1/4$ $m/s$ $\\$ (as $v$ = $1m/s$ from equation (1))

Now the two block have velocity $1/4$ $m.s$. at its mean poison.

$KE_{mass}$ = $(1/2)$$m^{|}v^{2} = (1/2) 4 x (1/4)^2 = (1/2) x (1/4). When the block are going to the extreme position, there will be only potential energy. \therefore PE = (1/2)$$k\delta^{2}$ = $(1/2)$ x $(1/4)$ where $\delta$ $\rightarrow$ new amplitude.

$\therefore$ $1/4$ = $100$ $\delta^{2}$ $\Rightarrow$ $\delta$ = $\sqrt{\frac{1}{400}}$ = $0.05$$m = 5cm. So amplitude = 5cm. 28. The left block in figure (12-E13) moves at a speed v towards the right block placed in equilibrium. All collisions to take place are elastic and the surfaces are frictionless. Show that the motions of the two blocks are periodic. Find the time period of these periodic motions. Neglect the widths of the blocks. When the block A moves with velocity 'V' and collides withe the block B, it transfers all energy to the \\ block B.(Because it is a elastic colosion).The block A will move a distance 'x' against the spring, again the block B will return to the original point and completes half of the oscillation. So, the time period of B is \frac{2\pi\sqrt{\frac{m}{k}}}{2} = \pi\sqrt{\frac{m}{k}} The block B collides with the block A and comes to rest at the point. \\ The block A again moves a further distance 'L' to return to its original position. \\ \therefore Time taken by the block to move from M \rightarrow N and N \rightarrow M is \frac{L}{V} + \frac{L}{V} = 2(\frac{L}{V}) \therefore So time period of the periodic motion is 2(\frac{L}{V}) + \pi\sqrt{\frac{m}{k}} 29. Find the time period of the motion of the particle shown in figure (12-E14). Neglect the small effect of the bend near the bottom. Let the time taken to travel AB and BC be t_1 and t_2 respectively For part AB,a_1 = g sin 45^0. s_1 = \frac{0.1}{sin 45^0} = 2m Let v = velocity at B \\ \therefore v^2 - u^2 = 2a_1s_1 \Rightarrow v^2 = 2 x g sin 45^0. s_1 = \frac{0.1}{sin 45^0} = 2 \Rightarrow v = \sqrt{2}m/s \therefore t_1 = \frac{v-u}{a_1} = \frac{\sqrt{2 - 0}}{\frac{g}{\sqrt2}} \therefore t_1 = \frac{v-u}{a_1} = \frac{\sqrt{2} - 0}{\frac{g}{\sqrt2}} = \frac{2}{g} = \frac{2}{10} = 0.2 sec Again for part BC , a_2 = -g sin 60^0 , u = \sqrt{2} , v = 0 \therefore t_1 = \frac{0 - \sqrt{2}}{-g(\frac{\sqrt{3}}{2})} = \frac{2\sqrt{2}}{\sqrt{3g}} = \frac{2 \times (1.414)}{(1.732) \times 10} = 0.165$$sec$.

So, time period = $2(t_1 + t_2)$ = $2($0.2$+$0.1555$)$ = $0.71$ $sec$

30.   All the surfaces shown in figure $(12-E15)$ are frictionless. The mass of the car is $M$, that of the block is $m$ and the spring has spring constant $k$. Initially, the car and the block are at rest and the spring is stretched through a length $x_0$, when the system is released. $\\$ (a) Find the amplitudes of the simple harmonic motion of the block and of the car as seen from the road. $\\$ (b) Find the time period(s) of the two simple harmonic motions.

Let the amplitude of oscillation of '$m$' and '$M$' be $x_1$ and $x_2$ respectively.

$a)$ From law of conservation of momentum,

$mX_1$ = $mX_2$ .....$(1)$ $[$ because only internal force are present $]$

Again, $(1/2)$ $kx_g^{2}$ = $(1/2)$ $k$ $(x_1 + x_2)^2$

$\therefore$ $x_0$ = $x_1$ + $x_2$ .....$(2)$

$[$ Block and mass oscillation in opposite direction. But $x$ $\rightarrow$ stretched part $]$

From equation $(1)$ and $(2)$

$\therefore$ $x_0$ = $x_1$ + $\frac{m}{M}$ $x_1$ = $(\frac{M+m}{M})$ $x_1$

$\therefore$ $x_1$ $\frac{mx_0}{M+m}$

So , $x_2$ = $x_0$ - $x_1$ = $x_0$ $[1-\frac{M}{M+m}]$ = $\frac{mx_0}{M+m}$ respectively.

$b)$ At any position , let the velocity be $v_1$ and $v_2$ respectively.

Here, $v_1$ = velocity of '$m$' with respect to $M$.

By energy method $\\$ Total Energy = Constant

$(1/2)$ $Mv^2$ + $(1/2)$ $m(V_1 - V_2)^2$ + $(1/2)$$k$$(X_1 + X_2)^2$ = Constant ...(i)

$[$ $v_1$ - $v_2$ = Absolute velocity of mass '$m$' as seen from the road.$]$

Again , from law of conservation of momentum.

$mv_2$ = $m(v_1- v_2)$ $\Rightarrow$ $(v_1- v_2)$ = $\frac{M}{m}$$v_2 ....(2) putting the above value in equation (1), we get \frac{1}{2}$$Mv_2^{2}$ + $\frac{1}{2}$$m\frac{M^2}{m^2}v_2^2 + \frac{1}{2}kx_2^2 (1+\frac{M}{m})^2 = constant \therefore M$$(1+\frac{M}{m})v_2$ + $k$$(1+\frac{M}{m})^2 x_2^2 = constant. \Rightarrow mv_2^2 + k$$(1+\frac{M}{m})^2$ $x_2^2$ = constant

Taking derivation of both sides,

$M$ x $2v_2$ $\frac{dv_2}{dt}$ + $k$$\frac{(M+m)}{m}-ex_2^2 \frac{dv_2}{dt} = 0 \Rightarrow ma_2 + k$$\frac{(M+m)}{m}$$x_2 = 0 [ because , v_2 = \frac{dx_2}{dt} ] \Rightarrow \frac{a_2}{X_2} = - \frac{k(M+m)}{Mm} = \omega^2 \therefore \omega = \sqrt{\frac{k(M+m)}{Mm}} So, Time period, T = 2\pi$$\sqrt{\frac{Mm}{k(M+m)}}$

31.   A uniform plate of mass $M$ stays horizontally and symmetrically on two wheels rotating in opposite directions $(figure 12-E16)$. The separation between the wheels is $L$. The friction coefficient between each wheel and the plate is $N$. Find the time period of oscillation of the plate if it is slightly displayed along its length and released

Let '$x$' be the displacement of the plank towards left.Now the center of gravity is also displaced through '$x$'

In displaced position

$R_1$ + $R_2$ = $mg$

Taking moment about $G$, we get

$R_1(\frac{\iota}{2})$ - $R_1x$ = $mg$ $\frac{\iota}{2}$ - $R_1x$ + $mgx$ - $R_1$$\frac{\iota}{2} So , R_1$$(\iota/2 - x)$ = $R_2$$(\iota/2 - x) = (mg - R_1)$$(\iota/2 - x)$

$\Rightarrow$ $R_1$$\frac{\iota}{2}-R_1x = mg \frac{\iota}{2} - R_1x + mgx - R_1$$\frac{\iota}{2}$

$\Rightarrow$ $R_1$$\frac{\iota}{2} + R_1$$\frac{\iota}{2}$ = $mg(x+\frac{\iota}{2})$

$\Rightarrow$ $R_1$$(\frac{\iota}{2} + \frac{\iota}{2}) = mg$$(\frac{2x+\iota}{2})$

$\Rightarrow$ $R_1\iota$ = $\frac{mg(2x+\iota)}{2}$

$\Rightarrow$ $R_1$ = $\frac{mg(2x+\iota)}{2\iota}$ ....$(2)$

Now $F_1$ = $\mu R_1$ = $\frac{\mu mg(\iota-2x)}{2\iota}$

Similarly $F_2$ = $\mu R_2$ = $\frac{\mu mg(\iota-2x)}{2\iota}$

Since, $F_1$ > $F_2$ $\Rightarrow$ $F_1$ - $F_2$ = $ma$ = $\frac{2\mu mg}{\iota}x$

$\Rightarrow$ $\frac{a}{x}$ = $\frac{2\mu g}{\iota}$ = $\omega^2$ $\Rightarrow$ $\omega$ = $\sqrt{\frac{2mu g}{\iota}}$

$\therefore$ Time period = $2\pi$ $\sqrt{\frac{\iota}{2rg}}$

32.   A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find, the length of a seconds pendulum at a place where $g$ = $\pi^2$m/s$^2$.

$T$ = $2$ $sec$.

$T$ = $2\pi$$\sqrt{\frac{\iota}{g}} \Rightarrow 2 = 2\pi$$\sqrt{\frac{\iota}{g}}$ $\Rightarrow$ $\frac{\iota}{g}$ = $\frac{1}{\pi^2}$ $\Rightarrow$ $\iota$ = $1$ $cm$ ( $\therefore$ $\pi^2$ $\approx$ $10$ $)$

33.   The angle made by the string of a simple pendulum with the vertical depends on time as $\theta$ = $\frac{\pi}{90}$ $sin$ $[(\pi s^{-1})t]$.Find the length of the pendulum if $g$ = $\pi^2$m/s$^2$.

From the equation.

$\theta$ = $\pi$ $sin$ $[(\pi sec^{-1})t]$

$\therefore$ $\omega$ = $\pi sec^{-1}$ $($ comparing with the equation of S.H.M$)$

$\Rightarrow$ $\frac{2\pi}{T}$ = $\pi$ $\Rightarrow$ $T$ = $2$ $sec$.

We know that $T$ = $2\pi$ $\sqrt{\frac{\iota}{g}}$ $\Rightarrow$ $2$ = $2$$\sqrt{\frac{\iota}{g}} \Rightarrow 1 = \sqrt{\frac{\iota}{g}} \Rightarrow \iota = 1$$m$.

$\therefore$ Length of the pendulum is $1$$m. 34. The pendulum of a certain clock has time period 2.04 s. How fast or slow does the clock run during 24 hours ? For the pendulum, \frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} Given that, T_1 =2sec ,g_1 = 9.8$$m/s$$^2 T_2 = \frac{24\times3600}{(\frac{24 \times 3600 - 24}{})} = 2 x \frac{3600}{3599} Now , \frac{g_2}{g_1} = ($$\frac{T_1}{T_2}$$)$$^2$

The pendulum of the clock has time period $2.04$$sec. Now, No. or oscillation in 1 day = \frac{24 \times 3600}{2} = 43200 But , in each oscillation it is slower by (2.04 - 2.00) = 0.04$$sec$

So in one day it is slower by,

= $43200$ x $(0.04)$ = $12$$sec = 28.8 min So, the clock runs 28.8 minutes slower in one day. 35. A pendulum clock giving correct time at a place where g = 9.800 m/s$$^2$ is taken to another place where it loses $24$ seconds during $24$ hours. Find the value of $g$ at this new place.

For the pendulum, $\frac{T_1}{T_2}$ = $\sqrt{\frac{g_2}{g_1}}$ $\\$ Given that, $T_1$ =$2sec$ ,$g_1$ = $9.8$$m/s$$^2$$\\ T_2 = \frac{24\times3600}{(\frac{24 \times 3600 - 24}{})} = 2 x \frac{3600}{3599}$$\\$ Now , $\frac{g_2}{g_1}$ = $($$\frac{T_1}{T_2}$$)$$^2$$\\$ $\therefore$ $g_2$ = $(9.8)$ $($ $\frac{3599}{3600}$$)$$^2$ = $9.795$$m/s$$^2$

36.   A simple pendulum is constructed by hanging a heavy ball by a $5.0$ $m$ long string. It undergoes small oscillations. $\\$(a) How many oscillations does it make per second ? $\\$(b) What will be the frequency if the system is taken on the moon where acceleration due to gravitation of the moon is $1.67$ $m/s$ $^2$.

$L$ = $5m$.

$a)$ $T$ = $2\pi$$\sqrt{\frac{\iota}{g}} = 2\pi$$\sqrt{0.5}$ = $2\pi$$(0.7) \\ \therefore In 2\pi$$(0.7)$$sec, the body completes 1 oscillation, \\ In 1 second, the body will complete \frac{1}{2\pi(0.7)} oscillation \\ \therefore f = \frac{1}{2\pi(0.7)} = \frac{10}{14\pi} = \frac{0.70}{\pi} times b) When it is taken to the moon T = 2\pi$$\frac{\iota}{g^{'}}$ $\\$ where $g^{'}$ $\rightarrow$ Acceleration in the moon.

= $2\pi$$\frac{5}{1.76} \therefore f = \frac{1}{T} = \frac{1}{2\pi}$$\sqrt{\frac{1.67}{5}}$ = $\frac{1}{2\pi}$$(0.577) = \frac{1}{2\pi\sqrt{3}}times. 37. The maximum tension in the string of an oscillating pendulum is double of the minimum tension. Find the angular amplitude. The tension in the pendulum is maximum at the mean position and minimum on the extreme position Here (1/2) mv^2 - 0 = mg \iota$$($$1 - cos$$\theta$$) v^2 = 2g \iota$$($$1 - cos$$\theta$$) Now, T_{max} = mg + 2 mg ($$1$ - $cos$$\theta$$)$ $\\$ [$T$ = $mg$ + ($mv^2$/ $\iota$)]

Again, $T_{min}$ = $mg$ $cos$$\theta. \Rightarrow mg + 2mg - 2mg cos$$2\theta$ = $2mg$ $cos$$\theta \Rightarrow 3mg = 4mg cos$$\theta$

$\Rightarrow$ $cos$$\theta = 3/4 \Rightarrow \theta = cos^{-1} (3/4) 38. A small block oscillates back and forth on a smooth concave surface of radius R (figure 12-E17). Find the time period of small oscillation. Given that, R = radius. \\ Let N = normal reaction. \\ Driving force F = mg sin$$\theta$.$\\$ Acceleration = $a$ = $mg$ $sin$$\theta As, sin$$\theta$ is very small, $sin$$\theta \rightarrow \theta therefore Acceleration a = g\theta Let ‘x’ be the displacement from the mean position of the body, \therefore \theta = \frac{x}{R} \Rightarrow a = g\theta = g(x/R) \Rightarrow (a/x) = (g/R) So the body makes S.H.M. \therefore T = 2\pi$$\sqrt{\frac{Displacement}{Acceleration}}$ = $2\pi$$\sqrt{\frac{x}{gx/R}} = 2\pi$$\sqrt{\frac{R}{g}}$

39.   A spherical ball of mass $m$ and radius $r$ rolls without slipping on a rough concave surface of large radius $R$. It makes small oscillations about the lowest point, Find the time period.

Let the angular velocity of the system about the point os suspension at any time be '$\omega$'

So, $v_c$ = $(R-r)$$\omega Again v_c = r$$\omega_1$ [where ,$\omega_1$ = rotational velocity of the sphere]

$\omega_1$ = $\frac{v_c}{r}$ = ($\frac{R + (-r)}{r}$)$\omega$ ....$(1)$

By Energy method, Total energy in SHM is constant.

So, $mg$$(R-r)(1-cos\theta) + (1/2)$$mv_c^{2}$+$(1/2)$ $|$$\omega_1^{2} = constant \therefore mg$$(R-r)(1-cos\theta)$ + $(1/2)$$m(R-r)^2 \omega^2 + (1/2) mr^2$$(\frac{R-r}{r})^2$$\omega^2 = constant \Rightarrow g$$(R-r)(1-cos\theta)$ + $(R-r)^2$ $\omega^2$ $[$$\frac{1}{2} + \frac{1}{5}$$]$ = constant

Taking derivative, $g(R-r)$ $sin$ $\theta$ $\frac{d\theta}{dt}$ = $\frac{7}{10}$ $(R-r)^2$$2\omega$$\frac{d\omega}{dt}$

$\Rightarrow$ $g$ $sin$ $\theta$ = $2$ x $\frac{7}{10}$ $(R-r)$$\alpha \Rightarrow g sin \theta = \frac{7}{5} (R-r)$$\alpha$

$\Rightarrow$ = $\frac{5gsin\theta}{7(R-r)}$ = $\frac{5g\theta}{7(R-r)}$

$\therefore$ $\frac{\alpha}{\theta}$ = $\omega$$^2 = \frac{5g\theta}{7(R-r)} = constant So the motion is S.H.M. Again \omega = \omega \sqrt{\frac{5g\theta}{7(R-r)}} \Rightarrow T = 2\pi \sqrt{\frac{7(R-r)}{5g}} 40. A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep. Calculate the time period of the pendulum there. Radius of the earth = 6400 km Length of the pendulum = 40$$cm$ = $0.4$. $\\$ Let acceleration due to gravity be g at the depth of $1600km$.

$\therefore$ $gd$ = $g(1-d/R)$ = $9.8$ $(1-\frac{1600}{6400})$ = $9.8$ $(1-\frac{1}{4})$ =$9.8$ x $\frac{3}{4}$ = $7.35$$m/s$$^2$

$\therefore$ Time period $T^{'}$ = $2\pi$$sqrt{\frac{l}{g\delta}} = 2\pi$$\sqrt{\frac{0.4}{7.35}}$ = $2\pi$ $\sqrt{0.054}$ = $2\pi$ x $0.023$ = $2$ x $3.14$ x $0.23$ = $1.465$ $\approx$ $1.47$$sec. 41. Assume that a tunnel is dug across the earth (radius = R) passing through its centre. Find the time a particle takes to cover the length of the tunnel if \\ (a) it is projected into the tunnel with a speed of \sqrt{gR} \\ (b) it is released from a height R above the tunnel \\ (c) it is thrown vertically upward along the length of tunnel with a speed of \sqrt{gR}. Let M be the total mass of the earth.\\ At any position x, \therefore \frac{M^{'}}{M} = \frac{\rho\times(\frac{4}{3})\pi \times x^{3}}{\rho \times(\frac{4}{3})\pi \times R^{3}} = \frac{x^3}{R^3} \Rightarrow M^{'} = \frac{Mx^3}{R^3} So force on the particle is given by, \therefore F_X = \frac{GM^{'}m}{x^2} = \frac{GMm}{R^3} ......(1) So, acceleration of the mass 'M' at the position is given by, a_x = \frac{GM}{R^2}$$x$ $\Rightarrow$ $\frac{a_x}{x}$ = $w^2$ = $\frac{GM}{R^3}$ = $\frac{g}{R}$ ($\therefore$ $\frac{Gm}{R^2}$)

So, $T$ = $2\pi$$\sqrt\frac{R}{g} = Time period of oscillation. a) Now, using velocity - displacement equation. V = \omega \sqrt{(A^2 - R^2)} [Where, A = amplitude] Given when, y = R, v = \sqrt{gR} , \omega = \sqrt{\frac{g}{R}} \Rightarrow \sqrt{gR} = \sqrt{\frac{g}{R}} \sqrt{(A^2 - R^2)} [because \omega = \sqrt{\frac{g}{R}}] \Rightarrow R^2 A^2 - R^2 \Rightarrow A = \sqrt{2}R [Now, the phase of the particle at the point P is greater than \pi/2 but less than \pi and at Q is greater than \pi but less than 3\pi /2. Let the times taken by the particle to reach the positions P and Q be t_1 & t_2 respectively, then using displacement time equation] y = r sin \omega t We have ,R = \sqrt{2} R sin \omega t_1 \\ \Rightarrow$$\omega t_1$ = $3\pi /4$

& $-R$ = $\sqrt{2}$ $R$ $sin$ $\omega t_2$ $\\$ $\Rightarrow$$\omega t_2 = 5\pi /4 So , \omega (t_1 - t_2) = \pi / 2 \Rightarrow t_1 - t_2 = \frac{\pi}{2\omega} = \frac{\pi}{2\sqrt{(R/g)}} Time taken by the particle to travel from P to Q is t2 – t1 \frac{\pi}{2\sqrt{(R/g)}} sec. b) When the body is dropped from a height R, then applying conservation of energy, change in P.E. = gain in K.E. \Rightarrow \frac{GMm}{R} - \frac{GMm}{2R} = \frac{1}{2} mv^2 \Rightarrow v = \sqrt{gR} Since, the velocity is same at P, as in part (a) the body will take same time to travel PQ. c) When the body is projected vertically upward from P with a velocity \sqrt{gR}, its velocity will be Zero at the highest point.\\ The velocity of the body, when reaches P, again will be v = \sqrt{gR} , hence, the body will take same\\ time \frac{\pi}{2\sqrt{(R/g)}} to travel PQ. 42. Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance R/2 from the earth's centre where R is the radius of the earth. The wall of the tunnel is frictionless.\\ (a) Find the gravitational force exerted by the earth on a particle of mass m placed in the tunnel at a distance x from the centre of the tunnel.\\ (b) Find the component of this force along the tunnel and perpendicular to the tunnel.\\ (c) Find the normal force exerted by the wall on the particle.\\ (d) Find the resultant force on the particle. \\ (e) Show that the motion of the particle in the tunnel is simple harmonic and find the time period. M = 4/3 \pi R^3 \rho. M^1 = 4/3 \pi x^3 _1 \rho. m^1 = (\frac{M}{R^3})x^3_1 a) F = Gravitational force exerted by the earth on the particle of mass ‘x’ is, F4 = \frac{Gm^1 m}{x_1^2} = \frac{GMm}{R^3} \frac{x^3_1}{x^2_1} = \frac{GMm}{R^3}$$x_1$ = $\frac{GMm}{R^3}$ $\sqrt{x^2 + (\frac{R^2}{4})}$

$b)$ $F_y$ = $F$ $cos$ $\theta$ = $\frac{GMm}{R^3}$$\frac{x}{X_1} = \frac{GMm}{R^3} F_x = F sin \theta = \frac{GMmx_1}{R^3} \frac{R}{2X_1} = \frac{GMm}{2R^2} c) F_x = \frac{GMm}{2R^2} { since Normal force exerted by the wall N = F_x] d) Resultant force = \frac{GMmx}{R^3} e) Acceleration = \frac{Driving force}{mass} = \frac{GMmx}{R^3m} \frac{Gmx}{R^3} So, a \alpha x (The body makes SHM) \therefore \frac{a}{x} = W^2 = \frac{GM}{R^3} \Rightarrow w = \sqrt{\frac{GM}{R^3}} \Rightarrow T = 2 \pi$$\sqrt{\frac{R^3}{GM}}$

43.   A simple pendulum of length $l$ is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator $\\$(a) is going up with an acceleration $a_0$ $\\$(b) is going down with an acceleration $a_0$, and$\\$ (c) is moving with a uniform velocity.

Here driving force $F$ = $m(g+a_0)$ $sin$ $\theta$ ....(1)

Acceleration $a$ = $\frac{F}{m}$ = ($g+a_0$)$sin$$\theta = \frac{(g+a_0)x}{l} (Because when \theta is small sin \theta \rightarrow \theta = \frac{x}{l} \\ \therefore a = \frac{(g+a_0)x}{l}. \therefore acceleration is proportional to displacement.\\ So, the motion is SHM.\\ Now \omega^2 = \frac{(g+a_0)}{l}. \therefore T = 2\pi \sqrt{\frac{l}{g+a_0}} b) When the elevator is going downwards with acceleration a_0 Driving force = F = m (g-a_0) sin$$\theta$.

Acceleration = $(g-a_0)$ $sin$$\theta = \frac{(g-a_0)x}{l} = \omega^2x T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{l}{g-a_0}} c) When moving with uniform velocity a_0 = 0. For, the simple pendulum, driving force = \frac{mgx}{l} \Rightarrow a = \frac{gx}{l} = \Rightarrow \frac{x}{a} = \frac{l}{g} T = 2\pi \sqrt{\frac{displacement}{acceleration}} 2\pi \sqrt{\frac{l}{g}} 44. A simple pendulum of length 1 feet suspended from the ceiling of an elevator takes \pi/3 seconds to complete one oscillation. Find the acceleration of the elevator. Let the elevator be moving upward accelerating ‘a_0’ \\Here driving force F = m(g + a0) sin \theta$$\\$ Acceleration = $(g + a0)$ $sin$ $\theta$

= $(g + a0)$ $\theta$ $(sin theta \rightarrow \theta)$

$\frac{(g+a_0)x}{l}$ = $\omega^2x$

$T$ = $2\pi$ $\sqrt{\frac{l}{g+a_0}}$

Given that, $T$ = $\pi / 3$ $Sec$,$l$ = $1ft$ and $g$ = $32$ $ft/sec^2$

$\frac{\pi}{3}$ = $2\pi$ $\sqrt{\frac{1}{32+a_0}}$

$\frac{1}{9}$ = $4$($\frac{1}{32 + a}$)

$\Rightarrow$ $32 + a$ = $36$ $\\$ $\Rightarrow$ $a$ = $36$ - $32$ = $4$$ft/sec^2 45. A simple pendulum fixed in a car has a time period of 4 seconds when the car is moving uniformly on a horizontal road. When the accelerator is pressed, the time period changes to 3.99 seconds. Making an approximate analysis, find the acceleration of the car. When the car moving with uniform velocity T = 2\pi \sqrt{\frac{l}{g}} \Rightarrow 4 = 2\pi \sqrt{\frac{l}{g}} ...(1) When the car makes accelerated motion, let the acceleration be a_0 \\ T = 2\pi \sqrt{\frac{l}{g^2 + a_0^2}} \Rightarrow 3.99 = 2\pi \sqrt{\frac{l}{g^2 + a_0^2}} Now \frac{T}{T^{'}} = \frac{4}{3.99} = \frac{(g^2 + a_0^2)^{1/4}}{\sqrt{g}} Solving for 'a_0' we can get a_0 = g/10 ms^{-2} 46. A simple pendulum of length l is suspended from the ceiling of a car moving with a speed v on a circular horizontal road of radius r.\\ (a) Find the tension in the string when it is at rest with respect to the car. \\$$(b)$ Find the time period of small oscillation.

From the freebody diagram,

$T$ = $\sqrt{(mg^2 + (\frac{mv^2}{r^2})}$

= $\sqrt{g^2 + \frac{v^4}{r^2}}$ = ma , where $a$ = acceleration = $(g^2 + \frac{v^4}{r^2})^{1/2}$

The time period of small accellations is given by,

$T$ = $2\pi$ $sqrt{\frac{l}{g}}$ = $2\pi$ $\sqrt{\frac{l}{(g^2 + \frac{v^4}{r^2})^{1/2}}}$

47.   The ear-ring of a lady shown in figure $(12-E18)$ has a $3$ $cm$ long light suspension wire.$\\$ $(a)$Find the time period of small oscillations if the lady is standing on the ground.$\\$ $(b)$ The lady now sits in a merry-go-round moving at $4$ $m/s$ in a circle of radius $2$ $m$. $\\$Find the time period of small oscillations of the ear-ring.

$a)$ $l$ = $3cm$ = $0.03m$

$T$ = $2\pi$$\sqrt{\frac{l}{g}} = 2\pi$$\sqrt{\frac{0.03}{9.8}}$ = $0.34$ second.

$b)$ When the lady sets on the Merry-go-round the ear rings also experience centrepetal acceleration

$a$ = $\frac{v^2}{r}$ = $\frac{4^2}{2}$ = $8$ $m/s^2$

Resultant Acceleration $A$ = $\sqrt{g^2+a^2}$ = $\sqrt{100+64}$ = $12.8$ $m/s^2$

Time period $T$ = $2\pi$$\sqrt{\frac{l}{A}} = 2\pi \sqrt{\frac{0.03}{12.8}} = 0.30 second. 48. Find the timik period of small oscillations of the following systems.\\ (a) A metre stick suspended through the 20 cm mark. .\\(b) A ring of mass in and radius r suspended through a point on its periphery. .\\(c) A uniform square plate of edge a suspended through a corner. .\\(d) A unifrom disc of mass m and radius r suspended through a point r/2 away from the centre. A) M.I about the pt A = l = l_{C.G} + Mh^2 = \frac{Ml^2}{12} + MH_2 = \frac{Ml^2}{12} + m(0.3)^2 = M(\frac{\frac{1}{12}}{0.09}) = M (\frac{1+1.08}{12}) = M (\frac{2.08}{12}) \therefore T = 2\pi$$\sqrt{\frac{l}{mgl^{'}}}$ = $2\pi$$\frac{2.08m}{m\times 9.8 \times\ 0.3} (l^{'} = dis. between C.G and pt. of suspension) \\ \approx 1.52sec. b) Moment of in isertia about A l = l_{C.G} + mr^2 = mr^2 + mr^2 = 2$$mr^2$

$\therefore$ Time period = $2\pi$$\sqrt{\frac{l}{mgl}} = 2\pi$$\sqrt{\frac{2mr^2}{mgr}}$ = $2\pi$$\sqrt{\frac{2r}{g}} c) l_{zz} (comes) = m (\frac{a^2+a^2}{3}) = \frac{2ma^2}{3} In the \DeltaABC, l^2+l^2 = a^2 \therefore l = \frac{a}{\sqrt2} \therefore T = 2\pi \sqrt{\frac{l}{mgl}} = 2\pi \sqrt{\frac{2ma^2}{3mgl}} = 2\pi \sqrt{\frac{2a^2}{3ga\sqrt{2}}} = 2\pi \sqrt{\frac{\sqrt{8a}}{3g}} d) h = r/2 , l = r/2 = Dist. Between C.G and suspension point. M.L. about A, l = l_{C.G.} + Mh^2 = \frac{mc^2}{2} + n(\frac{r}{2})^2 = mr^2 (\frac{1}{2}+\frac{1}{4}) = \frac{3}{4}$$mr^2$

$\therefore$ $T$ = $2\pi$ $\sqrt{\frac{l}{mgl}}$ = $2\pi$ $\sqrt{\frac{3Mr^2}{4mgl}}$ = $2\pi$ $\sqrt{\frac{3R^2}{4g(\frac{r}{2})}}$ = $2\pi$ $\sqrt{\frac{3r}{2g}}$

49.   A uniform rod of length $l$ is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the rod.

Let $A$ $\rightarrow$ suspension of point.$\\$ $B$ $\rightarrow$ Center of Gravity.$\\$ $l^{'}$ = $l/2$, $h$ = $l/2$$\\ Moment of inertia about A is l = l_{C.G.} + mh^2 = \frac{ml^2}{12} + \frac{ml^2}{4} = \frac{ml^2}{3} \Rightarrow T = 2\pi \sqrt{\frac{l}{mg\frac{l}{2}}} = 2\pi$$\sqrt{\frac{2ml^2}{3mgl}}$ = $2\pi$$\sqrt{\frac{2l}{3g}} Let, the time period ‘T’ is equal to the time period of simple pendulum of length ‘x’. \therefore T = 2\pi$$\sqrt{\frac{x}{g}}$. So, $\frac{2l}{3g}$ = $\frac{x}{g}$ $\Rightarrow$ $x$ = $\frac{2l}{3}$ $\\$ $\therefore$ Length of the simple pendulum = $\frac{2l}{3}$

50.   A uniform disc of radius r is to be suspended through a small hole made in the disc. Find the minimum possible time period of the disc for small oscillations. What should be the distance of the hole from the centre for it to have minimum time period ?

Suppose that the point is ‘$x$’ distance from C.G.$\\$ Let $m$ = mass of the disc., Radius = $r$$\\ Here l = x M.I. about A = l_{C.G.} + mx^2 = mr^2/2+mx^2 = m(r^2/2+x^2) T 2\pi \sqrt{\frac{l}{mgl}} = 2\pi \sqrt{\frac{m\big(\frac{r^2}{2}+X^2\big)}{mgx}} = 2\pi \frac{m(r^2/2+x^2)}{2mgx} = 2\pi \sqrt{\frac{r^2+2x^2}{2gx}} ....(1) For T is minimum \frac{dt^2}{dx} = 0 \therefore \frac{d}{dx}$$T^2$ = $\frac{d}{dx}$ $\big(\frac{4\pi^2r^2}{2gx}+\frac{4\pi^2 2x^2}{2gx}\big)$

$\Rightarrow$ $\frac{2\pi^2r^2}{g}$ $\big(-\frac{1}{x^2}\big)$ + $\frac{4\pi^2}{g}$ = $0$

$\Rightarrow$ $-\frac{\pi^2r^2}{gx^2}$ + $\frac{2\pi^2}{g}$ = $0$

$\Rightarrow$ $-\frac{\pi^2r^2}{gx^2}$ + $\frac{2\pi^2}{g}$ $\Rightarrow$ $2x^2$ = $r^2$ $\Rightarrow$ $x$ = $\frac{r}{\sqrt2}$

So putting the value of equation (1)

$T$ = $2\pi$ $\sqrt{\frac{r^2+\big(\frac{r^2}{2}\big)}{2gx}}$ = $2\pi$ $\sqrt{\frac{2r^2}{2gx}}$ = $2\pi$ $\sqrt{\frac{r^2}{g(\frac{r}{\sqrt2})}}$ $2\pi$ $\sqrt{\frac{\sqrt{2}r^2}{gr}}$ = $2\pi$ $\sqrt{\frac{\sqrt{2r}}{g}}$

51.   A hollow sphere of radius $2$ $cm$ is attached to an $18$ $cm$ long thread to make a pendulum. Find the time period of oscillation of this pendulum. How does it differ from the time period calculated using the formula for a simple pendulum ?

According to Energy equation,

$mgl(1-cos\theta)$ + $(1/2)$ $l\omega^2$ = const.

$mg(0.2)(1-cos\theta)$ + $(1/2)$ $l\omega^2$ = $C$.$\\$ Again, l = $2/3$ $m(0.2)^2$ + $m(0.2)^2$ $\\$ = $m\big[\frac{0.008}{3}+0.04 \big]$

= $m\big(\frac{0.1208}{3} \big)m$. Where I $\rightarrow$ Moment of Inertia about the pt of suspension $A$ $\\$ From equation$\\$ Differenting and putting the value of I and $1$ is

$\frac{d}{dt}\big[mg(0.2)(1-cos\theta)+\frac{1}{2}\frac{0.1208}{3}m\omega^2\big]$ = $\frac{d}{dt}$(C)

$\Rightarrow$ $mg(0.2) sin\theta + \frac{d\theta}{dt} + \frac{1}{2}\big(\frac{0.1208}{3}\big)m2\omega$ $\frac{d\omega}{dt}$ = $0$

$\Rightarrow$ $2$ $sin \theta$ = $\frac{0.1208}{3}$ $\alpha$ $\big[because, g = 10 m/s^2 \big]$

$\Rightarrow$ $\frac{\alpha}{\theta}$ = $\frac{6}{0.1208}$ = $\omega^2$ = $58.36$ $\\$ $\Rightarrow$ $\omega$ = $7.3.$ So $T$ = $\frac{2\pi}{\omega}$ = $0.89sec$.

For simple pendulum $T$ = $2\pi$ $\sqrt{\frac{0.19}{10}}$ = $0.86sec$.

% more = $\frac{0.89 - 0.86}{0.89}$ = $0.3$. $\\$ $\therefore$ it is about $0.3$% larger than the calculated value.

52.   A closed circular wire hung on a nail in a wall undergoes small oscillations of amplitude 2°and time period 2 s. Find $\\$ (a) the radius of the circular wire, $\\$ (b) the speed of the particle farthest away from the point of suspension as it goes through its mean position, $\\$ (c) the acceleration of this particle as it goes through its mean position and $\\$ (d) the acceleration of this particle when it is at an extreme position.Take $g$ = $\pi^2$$m/s^2. (For a compound pendulum) a) T = 2\pi \sqrt{\frac{l}{mgl}} = 2\pi \sqrt{\frac{l}{mgr}} \\ The MI of the circular wire about the point of suspension is given by \\ \therefore I = mr^2 + mr^2 = 2 mr^2 is Moment of inertia about A. \therefore 2 2\pi \sqrt{2mr^2mgr} = 2\pi \sqrt{\frac{2r}{g}} \Rightarrow \frac{2r}{g} = \frac{1}{\pi^2} \Rightarrow r = \frac{g}{2\pi^2} = 0.5\pi = 50$$cm$.(Ans)

$b)$ $(1/2)$ $\omega^2$ - $0$ = $mgr$ $(1-cos\theta)$

$\Rightarrow$ $(1/2)$ $2mr^2$ - $\omega^2$ = $mgr$ $(1-cos2^0)$

$\Rightarrow$ $\omega^2$ = $g/r$ $(1-cos2^0)$

$\Rightarrow$ $\omega$ = $0.11$ $rad/sec$ $[$ putting the value of $g$ and $r$ $]$

$\Rightarrow$ $v$ = $\omega$ x $1r$ = $11$ $cm/sec$.

$c)$ Acceleration at the end position will be centripetal.$\\$ = $a_n$ = $\omega^2$ $(2r)$ = $(0.11)^2$ × $100$ = $1.2$ $cm/s^2$ $\\$ The direction of ‘$a_n$’ is towards the point of suspension.

$d)$ At the extreme position the centrepetal acceleration will be zero. But, the particle will still have $\\$ acceleration due to the SHM.$\\$ Because, $T$ = $2$ $sec$.$\\$ Angular frequency $\omega$ = $\frac{2\pi}{T}$ $(\pi = 3.14)$ $\\$ So, angular acceleration at the extreme position,

$\alpha$ = $\omega^2\theta$ = $\pi^2$ x $\frac{2\pi}{180}$ = $\frac{2\pi^3}{180}$ [$1^0$ = $\frac{\pi}{180}$ radious]

So, tangential acceleration = $\alpha(2r)$ = $\frac{2\pi^3}{180}$ x $100$ = $34$ $cm/s^2$.

53.   A uniform disc of mass $m$ and radius $r$ is suspended through a wire attached to its centre. If the time period of the torsional oscillations be $T$, what is the torsional constant of the wire.

M.I. of the centre of the disc. = $mr^2/2$

$T$ = $2\pi$ $\sqrt{\frac{l}{k}}$ = $2\pi$ $\sqrt{\frac{mr^2}{2k}}$ [where $K$ = Torsional constant]

$T^2$ = $4\pi^2$ $\frac{mr^2}{2k}$ = $2\pi^2$ $\frac{mr^2}{k}$

$\Rightarrow$ $2\pi^2$ $mr^2$ = $KT^2$ $\Rightarrow$ $K$ = $\frac{2mr^2\pi^2}{T^2}$

$\therefore$ Torsional constant $K$ = $\frac{2mr^2\pi^2}{T^2}$

54.   Two small balls, each of mass $m$ are connected by a light rigid rod of length $L$. The system is suspended from its centre by a thin wire of torsional constant $k$. The rod is rotated about the wire through an angle $\theta_0$, and released. Find the tension in the rod' as the system passes through the mean position.

The M.I of the two ball system

l = $2m$$(L/2)^2 = m L^2/2 At any position \theta during the oscillation, [fig-2] \\ Torque = k\theta So, work done during the displacement 0 to \theta_0, W = \int_a^b k\theta d\theta = k \theta_0^2/2 By work energy method,\\ (1/2) l \omega^2 - 0 = Work done = k \theta_0^2/2 \\ \therefore \omega^2 = \frac{k \theta{_0^2}}{2l} = \frac{k \theta{_0^2}}{mL} \\ Now, from the freebody diagram of the rod,\\ T^2 = \sqrt{(m\omega^2L^2)+(mg)^2} \\ = \sqrt{\big(m\frac{k\theta_0^2}{mL^2}\times L\big)^2 + m^2g^2} = \frac{k^2\theta_0^4}{L^2}+m^2g^2 55. A particle is subjected to two simple harmonic motions of same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motions is (a) 0°, (b) 60°, (c) 90°. The particle is subjected to two SHMs of same time period in the same direction/ Given, r1 = 3cm, r2 = 4cm and \phi = phase difference. Resultant amplitude = R = \sqrt{r^2_1 + r^2_2 +2r_1r_2cos\phi} a) When \phi = 0^0, \\ R = \sqrt{(3^2+ 4^2+2\times 3 \times 4 cos 0^0)} = 7cm b) When \phi = 60^0, \\ R = \sqrt{(3^2+ 4^2+2\times 3 \times 4 cos 60^0)} = 6.1cm \\ c) When \phi = 90^0, \\ R = \sqrt{(3^2+ 4^2+2\times 3 \times 4 cos 90^0)} = 5cm 56. Three simple harmonic motions of equal amplitudes A and equal time periods in the same direction combine. The phase of the second motion is 60° ahead of the first and the phase of the third motion is 60° ahead of the second. Find the amplitude of the resultant motion. Three SHMs of equal amplitudes ‘A’ and equal time periods in the same dirction combine.\\ The vectors representing the three SHMs are shown it the figure. \\ Using vector method,\\ Resultant amplitude = Vector sum of the three vectors \\ = A +A cos 60^0 + A cos 60^0 = A + A/2 + A/2 = 2A \\ So the amplitude of the resultant motion is 2A. 57. A particle is subjected to two simple harmonic motions given by \\ x_1 = 2.0 sin(100 t) and x_2 = 2.0 sin(120 \pi t + \pi /3) where x is in centimeter and t in second. Find the displacement of the particle at \\(a) t = 0.0125,\\ (b) t = 0.025. x_1 = 2 sin 100 \pi t \\ x_2 = W sin (120 \pi t + \pi/3) \\ So, resultant displacement is given by,\\ x = x_1 + x_2 = 2 [$$sin$ $(100\pi t)$ + $sin$ $(120\pi t + \pi/3)$$] a) At t = 0.0125s, \\ x = 2 [$$sin$ $(100\pi \times 0.0125)$ + $sin$ $(120\pi \times 0.0125 + \pi/3)$$] \\ = 2 [ sin 5\pi/4 + sin (3\pi/2 + \pi/3)$$]$ $\\$ = $2$ $[(-0.707)+(-0.5)]$ = $-2.41$$cm. b) At t = 0.025s, \\ x = 2 [$$sin$ $(100\pi \times 0.025)$ + $sin$ $(120\pi \times 0.025 + \pi/3)$$] \\ = 2 [ sin 5\pi/2 + sin (3\pi + \pi/3)$$]$ $\\$ = $2$ $[1+(-0.8666)]$ = $0.27$$cm. 58. A particle is subjected to two simple harmonic motions, one along the X-axis and the other on a line making an angle of 45° with the X-axis, The two motions are given by x = x_0 sin$$\omega$$t and s = s_0 sin$$\omega$$t Find the amplitude of the resultant motion. The particle is subjected to two simple harmonic motions represented by,\\ x = x_0 sin wt \\ s = s_0 sin wt \\ and, angle between two motions = \theta = 45° \\ \therefore Resultant motion will be given by, R = \sqrt{(x^2+s^2+2xs cos 45^0)} \\ = \sqrt{{x_0^2 sin^2 wt + s_0^2 sin^2wt+2x_0s)0sin^2wtx(1/\sqrt2)}} \\ = [x+0^2 + s_0^2 = \sqrt2 x_0s_0]$$^{1/2}$ $sin$ $wt$ $\\$ $\therefore$ Resultant amplitude = $[x+0^2 + s_0^2 = \sqrt2 x_0s_0]$$^{1/2}$