Simple Harmonic Motion (SHM) MCQ Questions & Answers in Oscillation and Mechanical Waves | Physics

$$\eqalign{ & {E_1} = \frac{1}{2}k{x^2},{E_2} = \frac{1}{2}k{y^2}, \cr & E = \frac{1}{2}k{\left( {x + y} \right)^2} = \frac{1}{2}k{x^2} + \frac{1}{2}k{y^2} + kxy \cr & = {E_1} + {E_2} + 2\sqrt {{E_1}{E_2}} = 2 + 8 + 2\sqrt {16} = 18J \cr} $$

Time period of body executing $$SHM$$  is given by
$$T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{x}{a}} \,\,......\left( {\text{i}} \right)$$
where, $$x$$ is displacement of the particle and $$a$$ is acceleration of the particle.
From Eq. (i)
$$\eqalign{ & \omega = \sqrt {\frac{a}{x}} \,\,{\text{or}}\,\,{\omega ^2} = \frac{a}{x} \cr & {\text{Here,}}\,\,a = 2.0\,m/{s^2} \cr & x = 0.02\,m \cr & \therefore {\omega ^2} = \frac{{2.0}}{{0.02}} \cr & {\text{or}}\,\,{\omega ^2} = 100 \cr & {\text{or}}\,\,\omega = 10\,rad/s \cr} $$

Time period of spring-mass system, is given by
$$\eqalign{ & T = 2\pi \sqrt {\left( {\frac{{{\text{displacement}}}}{{{\text{acceleration}}}}} \right)} \cr & \therefore {\text{Frequency, }}n = \frac{1}{T} = \frac{1}{{2\pi }}\sqrt {\frac{{{\text{acceleration}}}}{{{\text{displacement}}}}} \cr & n = \frac{1}{{2\pi }}\sqrt {\frac{g}{l}} \,......\left( {\text{i}} \right) \cr} $$
In case of vertical spring-mass system, in equilibrium position
$$kl = mg \Rightarrow \frac{g}{l} = \frac{k}{m}$$
where,
$$l =$$  extension in the spring and
$$m =$$  mass of the suspended body
$$k =$$  spring constant or force constant of spring.
$$\therefore $$ From Eq. (i), we have
$$n = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}} \,\,{\text{or}}\,\,n \propto \frac{1}{{\sqrt m }}\,\,{\text{or}}\,\,\frac{{{n_1}}}{{{n_2}}} = \sqrt {\frac{{{m_2}}}{{{m_1}}}} $$
but $${m_1} = m,\,{m_2} = 4m,{n_1} = n\left( {{\text{given}}} \right)$$
$$\therefore \frac{n}{{{n_2}}} = \sqrt {\frac{{4m}}{m}} = 2\,\,{\text{or}}\,\,{n_2} = \frac{n}{2}$$
Alternative
As we know that
$$\eqalign{ & T = 2\pi \sqrt {\frac{m}{k}} \,\,\left( {{\text{for spring mass system}}} \right) \cr & n = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}} \cr} $$
So, for two different masses suspended with same spring.
$$\eqalign{ & {n_1} = \frac{1}{{2\pi }}\sqrt {\frac{k}{{{m_1}}}} \,\,\left[ {k\,{\text{is}}\,{\text{same}}\,{\text{for}}\,{\text{both}}\,{\text{the}}\,{\text{cases}}\,{\text{as}}\,{\text{spring}}\,{\text{is}}\,{\text{same}}} \right] \cr & {n_2} = \frac{1}{{2\pi }}\sqrt {\frac{k}{{{m_2}}}} \cr & {\text{so,}}\,\,\frac{{{n_1}}}{{{n_2}}} = \sqrt {\frac{{{m_2}}}{{{m_1}}}} \cr & {\text{here,}}\,\,{m_2} = 4{m_1} \cr & {\text{so,}}\,\,\frac{{{n_1}}}{{{n_2}}} = \sqrt {\frac{{4{m_1}}}{{{m_1}}}} = \frac{2}{1} \cr & \Rightarrow {n_1} = 2{n_2} \cr & \Rightarrow {n_2} = \frac{{{n_1}}}{2} = \frac{n}{2}\,\,\left[ {{n_1} = n} \right] \cr} $$

Both the bodies oscillate in simple harmonic motion, for which the maximum velocities will be
Given that $${v_1} = {v_2} \Rightarrow {a_1}{\omega _1} = {a_2}{\omega _2}$$
$$\eqalign{ & \therefore {a_1} \times \frac{{2\pi }}{{{T_1}}} = {a_2} \times \frac{{2\pi }}{{{T_2}}} \cr & \Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{{{T_1}}}{{{T_2}}} = \frac{{2\pi \sqrt {\frac{m}{{{k_1}}}} }}{{2\pi \sqrt {\frac{m}{{{k_2}}}} }} = \sqrt {\frac{{{k_2}}}{{{k_1}}}} \cr} $$

$$\eqalign{ & T = \frac{{2V\sin \theta }}{g}\, \cr & \therefore 1 = \frac{{2V\sin {{45}^ \circ }}}{g} \cr & \,\therefore \upsilon = \sqrt {50} m{s^{ - 1}} \cr} $$

$$\eqalign{ & {\text{Here,}} \cr & \,x = {x_0}\cos \left( {\omega t - \frac{\pi }{4}} \right) \cr & \therefore {\text{Velocity,}}\,v = \frac{{dx}}{{dt}} = - {x_0}\omega \sin \left( {\omega t - \frac{\pi }{4}} \right) \cr & {\text{Acceleration,}} \cr & a = \frac{{dv}}{{dt}} = - {x_0}{\omega ^2}\cos \left( {\omega t - \frac{\pi }{4}} \right) \cr & = {x_0}{\omega ^2}\cos \left[ {\pi + \left( {\omega t - \frac{\pi }{4}} \right)} \right] \cr & = {x_0}{\omega ^2}\cos \left( {\omega t + \frac{{3\pi }}{4}} \right)\,......\left( 1 \right) \cr & {\text{Acceleration,}}\,a = A\cos \left( {\omega t + \delta } \right)......\left( 2 \right) \cr} $$
Comparing the two equations, we get
$$A = {x_0}{\omega ^2}\,{\text{and}}\,\delta = \frac{{3\pi }}{4}$$

For the pendulum to be again in the same phase, there should be difference of one complete oscillation.
If smaller pendulum completes $$n$$ oscillations the larger pendulum will complete $$\left( {n - 1} \right)$$  oscillations, so Time period of $$n$$ oscillations of first = Time period of $$\left( {n - 1} \right)$$  oscillations of second
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\,n{T_1} = \left( {n - 1} \right){T_2} \cr & {\text{or}}\,\,n2\pi \sqrt {\frac{{{l_1}}}{g}} = \left( {n - 1} \right)2\pi \sqrt {\frac{{{l_2}}}{g}} \cr & {\text{or}}\,\,n\sqrt {{l_1}} = \left( {n - 1} \right)\sqrt {{l_2}} \cr & {\text{or}}\,\,\frac{n}{{n - 1}} = \sqrt {\frac{{{l_2}}}{{{l_1}}}} = \sqrt {\frac{{2.0}}{{0.5}}} \cr & {\text{or}}\,\,\frac{n}{{n - 1}} = 2 \cr & {\text{or}}\,\,n = 2\,n - 2 \cr & \therefore n = 2 \cr} $$

Distance of c.m from $$\frac{m}{2}$$
$$\eqalign{ & = \frac{{\frac{m}{2} \times 0 + m \times l}}{{\frac{m}{2} + m}} = \frac{{2l}}{3} \cr & {I_{cm}} = \frac{m}{2}{\left( {\frac{{2l}}{3}} \right)^2} + m{\left( {\frac{l}{3}} \right)^2} = \frac{1}{3}m{l^2} \cr} $$
At the mean position
$$\eqalign{ & \frac{1}{2}{I_{cm}}{\omega ^2} = \frac{1}{2}k\theta _0^2 \cr & \therefore {\omega ^2} = \frac{k}{{{I_{cm}}}}\theta _0^2 \cr & {\omega ^2} = \frac{{3k}}{{m{l^2}}}\theta _0^2 \cr} $$
As we know, $$\omega = \sqrt {\frac{k}{{{I_{cm}}}}} $$

Tension in the rod when it passes through the mean position,
$$ = m{\omega ^2}\frac{l}{3} = m\left[ {\frac{{3k}}{{m{l^2}}}\theta _0^2} \right]\frac{l}{3} = \frac{{k\theta _0^2}}{l}$$

$$\eqalign{ & P.E.,\,V = \frac{1}{2}m{\omega ^2}{x^2}\,{\text{and}}\,K.E.,\,T = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right) \cr & \therefore \frac{T}{V} = \frac{{{a^2} - {x^2}}}{{{x^2}}} \cr} $$

The restoring torque (for small $$\theta $$)

$$\eqalign{ & {\tau _{{\text{rest}}}} = - \left[ {\frac{{kL\theta }}{2} \times \frac{L}{2}} \right] \times 2 = \frac{{k{L^2}}}{2}\left( { - \theta } \right) \cr & \therefore \alpha = \frac{{{\tau _{{\text{rest}}}}}}{I} = \frac{{\frac{{k{L^2}}}{2}}}{{\frac{{M{L^2}}}{{12}}}}\left( { - \theta } \right) = \frac{{6k}}{M}\left( { - \theta } \right) \cr & \therefore T = 2\pi \sqrt {\frac{M}{{6k}}} . \cr} $$