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

Magnitude of velocity of particle when it is at displacement $$x$$ from mean position
$$ = \omega \sqrt {{A^2} - {x^2}} $$
Also, magnitude of acceleration of particle in $$SHM$$
$$ = {\omega ^2}x$$
Given, when $$x = 2\,cm$$
$$\eqalign{ & \left| v \right| = \left| a \right| \Rightarrow \omega \sqrt {{A^2} - {x^2}} = {\omega ^2}x \cr & \Rightarrow \omega = \frac{{\sqrt {{A^2} - {x^2}} }}{x} = \frac{{\sqrt {9 - 4} }}{2} \cr & \Rightarrow {\text{Angular velocity}}\,\omega {\text{ = }}\frac{{\sqrt 5 }}{2} \cr} $$
∴ Time period of motion
$$T = \frac{{2\pi }}{\omega } = \frac{{4\pi }}{{\sqrt 5 }}s$$

In case of sustained force oscillations the amplitude of oscillations decreases linearly.

Time period of spring pendulum, $$T = 2\pi \sqrt {\frac{M}{k}} .$$
If mass is doubled then time period
$$T' = 2\pi \sqrt {\frac{{2M}}{k}} = \sqrt 2 T$$

$$\eqalign{ & {\text{Given}}\,\,t = 1s \cr & \therefore x = 5\cos \left( {2\pi + \frac{\pi }{4}} \right) = 5\cos \frac{\pi }{4} = \frac{5}{{\sqrt 2 }}m \cr & {\text{i}}{\text{.e}}{\text{., displacement at }}t = {\text{ }}1s{\text{ is}}\,\frac{5}{{\sqrt 2 }}m \cr} $$

Maximum velocity during SHM = $$A\omega = A\sqrt {\frac{k}{m}} $$
$$\left[ {\therefore \omega = \sqrt {\frac{k}{m}} } \right]$$
Here the maximum velocity is same and $$m$$ is also same
$$\therefore {A_1}\sqrt {{k_1}} = {A_2}\sqrt {{k_2}} \,\therefore \frac{{{A_1}}}{{{A_2}}} = \sqrt {\frac{{{k_2}}}{{{k_1}}}} $$

As we know, for particle undergoing $$SHM,$$
$$\eqalign{ & V = \root \omega \of {{A^2} - {X^2}} ;\,\,V_1^2 = {\omega ^2}\left( {{A^2} - x_1^2} \right) \cr & V_2^2 = {\omega ^2}\left( {{A^2} - x_2^2} \right) \cr} $$
Substracting we get,
$$\eqalign{ & \frac{{V_1^2}}{{{\omega ^2}}} + x_1^2 = \frac{{V_2^2}}{{{\omega ^2}}} + x_2^2 \cr & w = \sqrt {\frac{{V_1^2 - V_2^2}}{{x_2^2 - x_1^2}}} \Rightarrow T = 2\pi \sqrt {\frac{{x_2^2 - x_1^2}}{{V_1^2 - V_2^2}}} \cr} $$

Total energy of the particle executing $$SHM$$  at instant $$t$$ is given by
$$E = \frac{1}{2}m{\omega ^2}{a^2}\,......\left( {\text{i}} \right)$$
and kinetic energy of the particle at instant $$t$$ is given by
$$\eqalign{ & {E_K} = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)\,......\left( {{\text{ii}}} \right) \cr & {\text{when}}\,\,x = \frac{a}{2},{E_K} = \frac{1}{2}m{\omega ^2}\left( {{a^2} - \frac{{{a^2}}}{4}} \right) \cr & = \frac{1}{2}m{\omega ^2} \times \frac{3}{4}{a^2} \cr & {\text{or}}\,\,{E_K} = \frac{1}{2} \times \frac{3}{4}m{\omega ^2}{a^2}\,......\left( {{\text{iii}}} \right) \cr} $$
From Eqs. (i) and (iii)
$$\frac{{{E_K}}}{E} = \frac{3}{4} \Rightarrow {E_K} = \frac{3}{4}E$$

$$\eqalign{ & U = {U_0} + \alpha \,{x^2} \Rightarrow F = - \frac{{dU}}{{dx}} = - 2\alpha x \cr & a = \frac{F}{m} = - \frac{{2\alpha }}{m}x \cr & \Rightarrow {\omega ^2} = \frac{{2\alpha }}{m} \Rightarrow T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{m}{{2\alpha }}} \cr} $$

Springs on the left of the block are in series, hence their equivalent spring constant is
$${K_1} = \frac{{\left( {2k} \right)\left( {2k} \right)}}{{2k + 2k}} = k$$
Springs on the right of the block are in parallel, hence their equivalent spring constant is
$${k_2} = k + 2k = 3k$$
Now again both $${K_1}$$ and $${K_2}$$ are in parallel
$$\therefore {K_{{\text{eq}}}} = {k_1} + {k_2} = k + 3k = 4k$$
Hence, frequency is
$$f = \frac{1}{{2\pi }}\sqrt {\frac{{{K_{{\text{eq}}}}}}{M}} = \frac{1}{{2\pi }}\sqrt {\frac{{4k}}{M}} $$

When a particle is dropped from a height $$h$$ above the centre of tunnel.
(i) It will oscillate, through the earth to a height $$h$$ on both sides.
(ii) The motion of particle is periodic.
(iii) The motion of particle will not be $$S.H.M.$$