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

At mean position velocity is maximum
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.,}}\,\,{v_{\max }} = \omega a \Rightarrow \omega = \frac{{{v_{\max }}}}{a} = \frac{{16}}{4} = 4 \cr & \therefore v = \omega \sqrt {{a^2} - {y^2}} \Rightarrow 8\sqrt 3 = 4\sqrt {{4^2} - {y^2}} \cr & \Rightarrow 192 = 16\left( {16 - {y^2}} \right) \Rightarrow 12 = 16 - {y^2} \cr & \Rightarrow y = 2\,cm. \cr} $$

$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{figure,}}\,\,\frac{T}{2} = 0.3\pi , \cr & \therefore T = 0.6\,\pi \,\,{\text{and}}\,\,{\phi _0} = - \frac{\pi }{2}. \cr & {\text{Thus}}\,\,y = A\sin \left( {\omega t + {\phi _0}} \right) = 2\sin \left\{ {\frac{{2\pi }}{{0.6\pi }}t + \left( { - \frac{\pi }{2}} \right)} \right\} \cr & = 2\sin \left( {\frac{{10t}}{3} - \frac{\pi }{2}} \right). \cr} $$

Two springs of force constants $${{k_1}}$$ and $${{k_1}}$$ are in parallel. Hence
$$k' = {k_1} + {k_1} = 2\,{k_1}$$
The third spring $${k_2}$$ is in series with spring of force constant $${k'}.$$
$$\therefore \frac{1}{k} = \left[ {\frac{1}{{2{k_1}}} + \frac{1}{{{k_2}}}} \right]\,\,{\text{or}}\,\,k = {\left[ {\frac{1}{{2{k_1}}} + \frac{1}{{{k_2}}}} \right]^{ - 1}}$$

$$y = 5\sin \left( {t\pi + 4\pi } \right),$$    comparing it with standard equation
$$\eqalign{ & y = a\sin \left( {\omega t + \phi } \right) = a\sin \left( {\frac{{2\pi t}}{T} + \phi } \right) \cr & a = 5m\,\,{\text{and}}\,\,\frac{{2\pi t}}{T} = \pi t \Rightarrow T = 2\sec . \cr} $$

Let the spring constant of the original spring be $$k.$$
Then its time period $$T = 2\pi \sqrt {\frac{m}{k}} $$   where $$m$$ is the mass of oscillating body.
When the spring is cut into $$n$$ equal parts, the spring constant of one part becomes $$nk.$$ Therefore the new time period, $$T' = 2\pi \sqrt {\frac{m}{{nk}}} = \frac{T}{{\sqrt n }}$$

In case of damped vibration, amplitude at any instant $$t$$ is given by
$$a = {a_0}\,{e^{ - bt}}$$
where,
$${a_0} = $$  initial amplitude
$$b = $$  damping constant
Case I
$$\eqalign{ & t = 100\,T\,\,{\text{and}}\,\,a = \frac{{{a_0}}}{3} \cr & \therefore \frac{{{a_0}}}{3} = {a_0}\,{e^{ - b\left( {100\,T} \right)}} \cr & \Rightarrow {e^{ - 100\,bT}} = \frac{1}{3} \cr} $$
Case II
$$\eqalign{ & t = 200\,T \cr & a = {a_0}\,{e^{ - bt}} = {a_0}\,{e^{ - b\left( {200\,T} \right)}} \cr & = {a_0}{\left( {{e^{ - 100\,bT}}} \right)^2} \cr & = {a_0} \times {\left( {\frac{1}{3}} \right)^2} = \frac{{{a_0}}}{9} \cr} $$
Thus, after 200 oscillations, amplitude will become $$\frac{1}{9}$$ times.

We know that $$V = \root \omega \of {{A^2} - {x^2}} $$
Initially $$V = \root \omega \of {{A^2} - {{\left( {\frac{{2A}}{3}} \right)}^2}} $$
Finally $$3v = \root \omega \of {A{'^2} - {{\left( {\frac{{2A}}{3}} \right)}^2}} $$
Where $$A' = $$ final amplitude (Given at $$x = \frac{{2A}}{3},$$   velocity to trebled)
On dividing we get $$\frac{3}{1} = \frac{{\sqrt {A{'^2} - {{\left( {\frac{{2A}}{3}} \right)}^2}} }}{{\sqrt {{A^2} - {{\left( {\frac{{2A}}{3}} \right)}^2}} }}$$
$$9\left[ {{A^2} - \frac{{4{A^2}}}{9}} \right] = A{'^2} - \frac{{4{A^2}}}{9}\,\,\,\,\,\,\,\,\,\,\,\therefore A' = \frac{{7A}}{3}$$

$$\eqalign{ & {\text{Initially}}\,t = 0 \cr & x = a\cos \pi t = - a\cos \,{0^ \circ } = a \cr & {\text{Finally}}\,{\text{at}}\,\,t = 3 \cr & x = a\cos 3\pi = - 3a \cr & {\text{Total displacement}} = {\text{ }}2a \cr} $$

$$h$$ = Length of block immerged in water $$mg = {F_B}$$
$$\eqalign{ & \rho A\lg = {\rho _B}Ahg \cr & 650 \times A \times 54 \times {10^{ - 2}}g = 900 \times A \times hg \cr & \Rightarrow h = 0.39\,m = 39\,cm. \cr} $$

As, $$x = A\cos \omega t$$
$$\eqalign{ & \therefore v = \frac{{dx}}{{dt}} = - A\omega \sin \omega t\,......\left( {\text{i}} \right) \cr & {\text{and}}\,\,a = \frac{{{d^2}x}}{{d{t^2}}} = - A{\omega ^2}\cos \omega t\,......\left( {{\text{ii}}} \right) \cr} $$
We can find the correct graph by putting different values of $$t$$ in Eq. (ii).
$$\eqalign{ & {\text{At}}\,\,t = 0,a = - A{\omega ^2} \cr & {\text{At}}\,\,t = \frac{T}{4},a = - A{\omega ^2}\cos \left( {\frac{{2\pi }}{T} \times \frac{T}{4}} \right) = 0 \cr & {\text{At}}\,\,t = \frac{T}{2},a = - A{\omega ^2}\cos \left( {\frac{{2\pi }}{T} \times \frac{T}{2}} \right) \cr & = - A{\omega ^2}\cos \,\pi = + A{\omega ^2} \cr & {\text{At}}\,\,t = \frac{{3T}}{4},a = - A{\omega ^2}\cos \left( {\frac{{2\pi }}{T} \times \frac{{3T}}{4}} \right) = 0 \cr & {\text{At}}\,\,t = T,a = - A{\omega ^2}\cos \left( {\frac{{2\pi }}{T} \times T} \right) = - A{\omega ^2} \cr} $$
This condition is represented by graph in option (C).