Concept Of Physics Simple Harmonics Motion

H C Verma

Concept Of Physics

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 = $\frac{2\pi}{\omega}$ = $\frac{2}{10\sqrt5}$ = $\frac{2x3.14}{10x2.236}$ = $0.28$seconds.

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 $\approx$0.049$m$

$\therefore$ $r$ = 4.9 $cm$

2.   A particle executes simple harmonic motion with an amplitude of $10$ cm and time period $6$s. 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 $r$ = $10$cm

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 ($w$x$0$x$\phi$) = $10$ 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$$\pi$t = $tan^{-1}$ $0.75$) = $2.0$ $cos$ ($50$$\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}$

$\Rightarrow$b$4$ = $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 $5$x$16$ = $5$ 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 $\delta$ = $F/k$ = $10/100$ = $0.1$m = $10$ $cm$

$b)$ The below imparts a speed of $2m/s$ to block towards left.

$\therefore$ P.E. + K.E. = $1/2$ $k\delta^2$ + $1/2$ M$v^2$

= $(1/2)$ 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 $100(0.1+0.2)^{2}$ = $50$ 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]

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