Oscillations Practice Questions

At mean position, K.E. is maximum where as P.E. is minimum

In simple harmonic motion, starting from rest, At $t=0, x=A$ $x=A \cos \omega t \;\; . . .(i)$ When $t=\tau , x=A-a$ When $t=2 \tau , x=A-3 a$ From equation $(i)$ $A-a=A \cos \omega \tau \;\; . . .(i i)$ $A-3 a=A \cos 2 \omega \tau \;\; . . .(i i i)$ As $\cos 2 \omega \tau =2 \cos ^{2} \omega \tau -1 . . .(i v)$ From equation $(i i),(i i i)$ and $(i v)$ $\;\frac{A-3 A}{A}=2(\;\frac{A-a}{A})^{2}-1$ $\Rightarrow \;\frac{A-3 a}{A}=\;\frac{2 A^{2}+2 a^{2}-4 A a-A^{2}}{A^{2}}$ $\Rightarrow A^{2}-3 a A=A^{2}+2 a^{2}-4 A a$ $\Rightarrow 2 a^{2}=a A \Rightarrow A=2 a$ $\Rightarrow \;\frac{a}{A}=\;\frac{1}{2}$ Now, $A-a=A \cos \omega \tau$ $\Rightarrow \cos \omega \tau =\;\frac{A-a}{A} \Rightarrow \cos \omega \tau =\;\frac{1}{2}$ $\;\text{ or, }\; \;\frac{2 \pi }{T} \tau =\;\frac{\pi }{3} \Rightarrow T-6 \tau$

FBD of piston at equilibrium$\Rightarrow P_{\text{atm}}A+mg=P_0A$........(1)FBD of piston when piston is pushed down a distance x$P_{\text{atm}}+mg-(P_0+dP)a=m\frac{d^2x}{dt^2}$................(2)Process is adiabatic $\Rightarrow PV^\gamma =C\Rightarrow −dP={\gamma PdV}/V$Using 1, 2, 3 me get $f=1/{2\pi }\sqrt{\frac{A^2\gamma P_0}{MV_0}}$

$A=A_0e^{-kt}$ $\Rightarrow 0.9A_0=A_0e^{-5k}$and $\alpha A_0=A_0e^{-15k}$solving $\Rightarrow \alpha =0.79$

The equation of motion for the pendulum, suffering retardation$I \alpha =-m g(ℓ sin \;\theta )-m b v(ℓ)$ where $I=m ℓ^2$and $\alpha =d^2 \theta / \text{dt}^2$ $\therefore \;\frac{d^2 \theta }{d t^2}=-\;\frac{g}{ℓ} \tan \theta +\;\frac{b v}{ℓ}$on solving we get $\theta =\theta _0 e^{\;\frac{b t}{2} sin \;(\omega t+\varphi )}$According to questions $\;\frac{\theta _0}{e}=\theta _0 e^{\;\frac{-b r}{2}}$ $\therefore \tau =\;\frac{2}{b}$

For $X_0+A$ to be the maximum separation $y$ one body is at the mean position, the other should be at the extreme.

The net force becomes zero at the mean point. Therefore, linear momentum must be conserved.$\; \therefore M v_1=(M+m) v_2$ $\;M A_1 \sqrt{\;\frac{k}{M}}=(M+m) A_2 \sqrt{\;\frac{k}{m+M}}$ $\; \therefore (V=A \sqrt{\;\frac{k}{M}})$ $\;A_1 {\sqrt{M}}=A_2 {\sqrt{M+m}}$ $\; \therefore \;\frac{A_1}{A_2}=\sqrt{\;\frac{m+M}{M}}$

For an $S H M$, the acceleration $a=-\omega ^2 x$ where $\omega ^2$ is a constant. Therefore $\;\frac{a}{x}$ is a constant. The time period $T$ is also constant. Therefore $\;\frac{a T}{x}$ is a constant.

The two springs are in parallel.$f=\;\frac{1}{2 \pi } \sqrt{\;\frac{K_1+K_2}{m}}$........(i) $f^{'}=\;\frac{1}{2 \pi } \sqrt{\;\frac{4 K_1+4 K_2}{m}}$ $=\;\frac{1}{2 \pi } \sqrt{\;\frac{4 (K_1+4 K_2) }{m}}$ $=2 (\;\frac{1}{2 \pi } \sqrt{\;\frac{K_1+K_2}{m}})$ $=2 f \;\;$ from eqn. $(i)$

Here, $x=2 \times 10^{-2} \cos \pi t$ $\therefore v=\;\frac{d x}{d t}=2 \times 10^{-2} \pi \sin \;\pi t$For the first time, the speed to be maximum,$\; \sin \;\pi t=1$ or, $\sin \;\pi t= \sin \;\;\frac{\pi }{2}$ $\;\Rightarrow \pi t=\;\frac{\pi }{2} \;\; \;\text{ or, }\; \;\; t=\;\frac{1}{2}=0.5 {\ sec} .$

Here, $x=x_0 \cos (\omega t-\pi {/} 4)$ $\therefore$ Velocity, $v=\;\frac{d x}{d t}=-x_0 \omega \sin \; (\omega t-\;\frac{\pi }{4})$ Acceleration, $a=\;\frac{d v}{d t}=-x_0 \omega ^2 \cos (\omega t-\;\frac{\pi }{4})$ $=x_0 \omega ^2 \cos [\pi + (\omega t-\;\frac{\pi }{4}) ]$ $=x_0 \omega ^2 \cos (\omega t+\;\frac{3 \pi }{4})$ ....(i) Acceleration, $a=A \cos (\omega t+\delta )$...(ii) Comparing the two equations, we get$A=x_0 \omega ^2$and $\delta =\;\frac{3 \pi }{4} \;\text{. }\;$

The instantaneous kinetic energy of a particle executing $S . H . M$. is given by $K=\;\frac{1}{2} m a^2 \omega ^2 sin \;^2 \omega t$ $\therefore$ average $K . E .=< K>=< \;\frac{1}{2} m \omega ^2 a^2 sin \;^2 \omega t>$ $=\;\frac{1}{2} m \omega ^2 a^2< sin \;^2 \omega t>$ $=\;\frac{1}{2} m \omega ^2 a^2 (\;\frac{1}{2})$ $(.$ as $< sin \;^2 \theta >=\;\frac{1}{2})$ $=\;\frac{1}{4} m \omega ^2 a^2=\;\frac{1}{4} m a^2(2 \pi v)^2$ $(\omega =2 \pi v)$or, $< K>=\pi ^2 m a^2 v^2$

$K . E$. of a body undergoing $S H M$ is given by,$\;K . E .=\;\frac{1}{2} m a^2 \omega ^2 \cos ^2 \omega t$ $\;T . E .=\;\frac{1}{2} m a^2 \omega ^2$Given $K . E .=0.75 T . E$.$\;\Rightarrow 0.75=\cos ^2 \omega t \Rightarrow \omega t=\;\frac{\pi }{6}$ $\;\Rightarrow t=\;\frac{\pi }{6 \times \omega } \Rightarrow t=\;\frac{\pi \times 2}{6 \times 2 \pi } \Rightarrow t=\;\frac{1}{6} s$

Maximum velocity,$\;v_{\max }=a \omega , \;\; v_{\max }=a \times \;\frac{2 \pi }{T}$ $\;\Rightarrow T=\;\frac{2 \pi a}{v_{\max }}=\;\frac{2 \times 3.14 \times 7 \times 10^{-3}}{4.4} \approx 0.01 {\ s}$

Coin $A$ is moving in simple harmonic motion in vertical direction. Now we are assuming coin will leave contact with the platform when platform is at a distance of $x$ from the mean position which is also called amplitude.At distance $x$ the force acting on the coin is$m g-N=m \omega ^2 x$For coin to leave contact $N=0$ $\Rightarrow m g=m \omega ^2 x \Rightarrow x=\;\frac{g}{\omega ^2}$ $\therefore$ Option (B) is correct.

$\;y= sin \;^2 \omega t$ $\;=\;\frac{1-\cos 2 \omega t}{2}$ $\;=\;\frac{1}{2}-\;\frac{1}{2} \cos 2 \omega t$ $\therefore$ Angular speed $=2 \omega$ $\therefore$ Period $(T)=\;\frac{2 \pi }{\;\text{ angular speed }\;}=\;\frac{2 \pi }{2 \omega }=\;\frac{\pi }{\omega }$So it is a periodic function.As $y= sin \;^2 \omega t$ $\;\frac{d y}{d t}=2 \omega sin \;\omega t \cos \omega t=\omega sin \;2 \omega t$ $\;\frac{d^2 y}{d t^2}=2 \omega ^2 \cos 2 \omega t$ which is not proportional to $- {\text{y}}$.Hence it is is not SHM.

$\;v_1=\;\frac{d y_1}{d t}=0.1 \times 100 \pi \cos (100 \pi t+\;\frac{\pi }{3})$ $\;v_2=\;\frac{d y_2}{d t}=-0.1 \pi sin \;\pi t=0.1 \pi \cos (\pi t+\;\frac{\pi }{2})$ ∴ Phase diff. $=\varphi _1-\varphi _2=\;\frac{\pi }{3}-\;\frac{\pi }{2}=\;\frac{2 \pi -3 \pi }{6}=\;\frac{\pi }{6}$

Center of mass of combination of liquid and hollow portion (at position $ℓ$ ), first goes down (to $ℓ+\Delta ℓ$ ) and when total water is drained out, center of mass regain its original position (to $ℓ$ ), $T=2 \pi \sqrt{\;\frac{ℓ}{g}}$ $\therefore { }^{'} T^{'}$ first increases and then decreases to original value.