# 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 $F$ = $F_1+F_2$ = ($K_1+K_2$)$x$