The x-t graph of a particle undergoing SHM is shown in the figure. The acceleration of the particle at t=$1/3_s$ is

(A) $\frac{\sqrt{3}}{32} \pi^2 \mathrm{~cm} / \mathrm{s}^2$

(B) $\frac{-\pi^2}{32} \mathrm{~cm} / \mathrm{s}^2$

(C) $\frac{\pi^2}{32} \mathrm{~cm} / \mathrm{s}^2$

(D) $-\frac{\sqrt{3}}{32} \pi^2 \mathrm{~cm} / \mathrm{s}^2$

Simple harmonic motionequation: $x=A \sin (\omega t+\phi)$

From diagram: A=1, at t=0, x=0.

0 = $\sin\phi \Rightarrow \phi=0$

At T =8 sec, $\omega=\frac{2\pi}{T}=\frac{2\pi}{8}$

$x=\sin \left(\frac{\pi}{4} t\right)$

$v =\frac{d x}{d t}=\frac{d}{d t}\left(\sin \frac{\pi}{4} t\right)$

$=\frac{\pi}{4} \cos \frac{\pi}{4} t$

$a=\frac{d v}{d t}=\frac{d^2 x}{d t^2}=-\frac{\pi^2}{16} \sin \frac{\pi}{4} t$

$a\left( t=4 /3 s\right) =-\frac{\pi^2}{16} \sin (\frac{\pi}{4} \cdot \frac{4}{3}) = -\frac{\pi^2}{16} \sin \frac{\pi}{3} = -\frac{\pi^2}{16} \frac{\sqrt{3}}{2} = -\frac{\sqrt{3} \pi^2}{32} \mathrm{~cm} / \mathrm{s}^2$

$a\left(t=\frac{4}{3} s\right) = -\frac{\sqrt{3} \pi^2}{32} \mathrm{~cm} / \mathrm{s}^2$, Correct option : D

A point mass is subjected to two simultaneous sinusoidal displacements in $x$ direction, $x_1(t)=A \sin (\omega t)$ and $x_2(t)=A \sin \left(\omega t+\frac{2 \pi}{3}\right)$. Adding a third sinusoidal displacement $x_3(t)=B \sin (\omega t+\phi)$ brings the mass to complete rest. The value of $B$ and $\phi$ are (2011)

(A) $\sqrt{2} A, 3 \pi / 4$

(B) $A, 4 \pi / 3$

(C) $\sqrt{3} A, 5 \pi / 6$

(D) $A, \pi / 3$

$x_1 = A\sin\omega t$

$\vec{x}_1=(A, \quad \phi=0)$

$x_2=A \sin \left(\omega t+\frac{2 \pi}{3}\right)$

$\vec{x}_2=\left(A, \frac{2 \pi}{3}\right)$

$x_1+x_2+x_3=0$

$\Rightarrow \quad x_3=-\left(x_1+x_2\right)$

$=-A\left[\sin \omega t+\sin \left(\omega t+\frac{2 \pi}{3}\right)\right]$

$\sin C+\sin D=2 \sin \frac{C+D}{2} \cos \frac{C - D}{2}$

$x_3 =-A \sin \left(\omega t+\frac{\pi}{3}\right)$

$=A \sin \left(\omega t+\frac{\pi}{3}+\pi\right)$

$=A \sin \left(\omega t+\frac{4 \pi}{3}\right)$

$=B \sin (\omega t+\phi)$

B =A

$\phi = \frac{4\pi}{3}$

$\vec{x}_3=-\vec{x}=(A, \phi)$

$\phi=\pi+\frac{\pi}{3}=\frac{4}{3} \pi$

$\vec{x}_3=\left(A, \frac{4}{3} \pi\right)$

Correct option B

A block with mass $M$ is connected by a masses spring with spring constant $k$ to a rigid wall and mover without friction on a horizontal surface. The block oscillates with small amplitude $A$ about an equilibrium position $x_0$. Consider two cases: (1) when the block is at $x_0 ;$ and (2) when the block is at $x=x_0+A$. In both the cases, a particle with mass $m(<M)$ is softly placed on the block after which they stick to each other. Which of the following statement(s) is(are) true about the motion after the mass $m$ is placed on the mass $M$ ? (2016)

(A) The amplitude of oscillation in the first case changes by a factor of $\sqrt{\frac{M}{m+M}}$, whereas in the second case it remains unchanged.

(B) The final time period of oscillation in both the cases is same.

(C) The total energy decreases in both the cases.

(D) The instantaneous speed at $x_0$ of the combined masses decreases in both the cases.

$\omega=\sqrt{\frac{K}{M}}$

Time period $T=\frac{2 \pi}{\omega}$

$v=\omega A$

Total energy $E=\frac{1}{2} k A^2$

Case:1 $\omega_1 = \sqrt\frac{k}{M+m}$

Linear Momentum is conserved: $M v =(M+m) v_1$

$v_1=\frac{M}{M+m} \quad v =\frac{M}{M+m} \omega A \rightarrow(i)$

$v_1 <v$

Say, the new amplitude is A, $v_1=\omega_1 A_1$

$\Rightarrow A_1=\frac{v_1}{\omega_1}$

$\Rightarrow A_1=\sqrt\frac{M}{M+m} A \rightarrow (ii)$

Time period $T_1 =\frac{2\pi}{\omega_1} \rightarrow (iii)$

Total energy of the system

$E_1= \frac{1}{2} k A_1^2=\frac{1}{2} k A^2\left(\frac{M}{M+m}\right)$

$=\left(\frac{m}{M+m}\right) E \rightarrow (iv)$

Total energy decreases compared to the initial total energy.

Case:2 The angular frequency $\omega_2= \sqrt\frac{k}{M+m}$

If a man 'm' is put over the block due to conservation of linear momentum, velocity of the combined system just after m is put over M is also zero.

$A_2 = A$

Total Energy: $\frac{1}{2}kA_2 ^2=\frac{1}{2}kA ^2$

The total energy of the system remains unchanged in case 2.

Time period , $T_2=\frac{2\pi}{\omega_2}=\frac{2\pi}{\omega_1}= T_1$

Clearly, the final time period in bothe case is same.

(A) The amplitude of oscillation in the first case changes by a factor of $\sqrt{\frac{M}{m+M}}$, whereas in the second case it remains unchanged.

(B) The final time period of oscillation in both the cases is same.

(D) The instantaneous speed at $x_0$ of the combined masses decreases in both the cases.

Correct answer (A, B, D).

A uniform rod of length $L$ and mass $M$ is pivoted at the centre. Its two ends are attached to two springs of equal spring constants $k$. The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $\theta$ in one direction and released. The frequency of oscillation is (2009)

(A) $\frac{1}{2 \pi} \sqrt{\frac{2 k}{M}}$

(B) $\frac{1}{2 \pi} \sqrt{\frac{k}{M}}$

(C) $\frac{1}{2 \pi} \sqrt{\frac{6 k}{M }}$

(D) $\frac{1}{2 \pi} \sqrt{\frac{24 k}{M}}$

The restoring torque about O is in anticlockwise direction:

$\left(\frac{K}{2} L \theta\right)\left(\frac{L}{2}\right)+\left(\frac{K}{2} L \theta\right)\left(\frac{L}{2}\right)=\frac{1}{2} K L^2 \theta$

The moment of inertia of the rod about the oscillation axis is: $I=\frac{ML^2}{12}$

The angular acceleration of the rod is: $\alpha=-\frac{d^2\theta}{dt}$

$\tau=I \alpha$ = $\left(\frac{M L^2}{12}\right) \frac{d^2 \theta}{d t^2}=-\frac{1}{2} K L^2 \theta$

$\Rightarrow \quad \frac{d^2 \theta}{d t^2}=-\left(\frac{6 K}{M}\right) \theta=-\omega^2 \theta$

$\omega=\sqrt{\frac{6 K}{M}}$ = $\nu=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{6 K}{M}}$

Paragraph type question

Phase space diagrams are useful tools in analysing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one dimension.

For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is $x(t)$ versus $p(t)$ curve in this plane.

Say, the ball is thrown up with an initial velocity $v_0$ and has mass m.

$v^2=v_0^2-2gx \quad p=mv$

$p^2=m^2 v_0^2-2 m^2 g x$

$p= \pm \sqrt{m^2 v_0^2-2 m^2 g x}$

At x=0, p=$mv_0$ When the ball goes up

At x=0, p=-$mv_0$ When the ball comes back

At the maximum height, $x=\frac{v_0^2}{2g}$

The momentum becomes zero.

The phase space diagram for SHM is a circle centered at the origin. In the figure, two circles represent the same oscillator but for different initial conditions, and $E_1$ and $E_2$ are total mechanical energies, respectively. Then,

A) $E_1=\sqrt{2} E_2$

(B) $E_1=2 E_2$

(C) $E_1=4 E_2$

(D) $E_1=16 E_2$

$E=\frac{1}{2} m \omega^2 A^2$

$E_1=\frac{1}{2} m \omega^2(2 a)^2$

$E_2=\frac{1}{2} m \omega^2 a^2$

$\frac{E_1}{E_2}=4 \Rightarrow E_1=4 E_2$

$x=A \cos \omega t$

$p=m \frac{d x}{d t}=-m A \omega \sin \omega t$

When the system is submerged in water the dumping cause continuous reduction of amplitude.

A simple pendulum has time period $T_1$. The point of suspension is now mored upwards according to the relation $y=kt^2,\left(k=1 \frac{m}{s^2}\right)$ where $y$ is the vertical displacement. The time period now becomes $T_2$. The ratio of $T_1^2/T_2^2$ [Take $g=10 \frac{m}{s^2}$] (2005)

(A) $6 / 5$

(B) $5 / 6$

(C) 1

(D) $4 / 5$

$T_1=2 \pi \sqrt{\frac{L}{g}} \quad \rightarrow(i)$

$y=kt^2$

$\frac{d^2 y}{d t^2}=2 k=2 \mathrm{~m} / \mathrm{s}^2$

$a=2 \mathrm{~m} / \mathrm{s}^2$

The acceleration of the pendulum with respect to the Point of suspension O is

$a+g=12 \mathrm{~m} / \mathrm{s}^2$

$T_2=2 \pi \sqrt{\frac{L}{a+g}} \rightarrow(i i)$

$\frac{T_1^2}{T_2^2}=\frac{a+g}{g}=\frac{12}{10}=\frac{6}{5}$

A particle of mass $m$ is executing oscillations about the origin on the $x$-axis. Its potential energy is $U(x)=k|x|^3$, where $k$ is a positive constant. If the amplitude of oscillation is $a$, then its time period $T$ is

(1998)

(A) proportional to $1 / \sqrt{a}$

(B) independent of $a$

(C) proportional to $\sqrt{a}$

(D) proportional to $a^{3 / 2}$

$v(x)=\frac{1}{2} k x^2$

This is not exactly harmonic oscillator Because in H.O. $U(x)=k|x|^2$

At x=a, $U = k a^3$

$k=\frac{1}{2} m v^2=0$

$E=v+k=k a^3$

$0<x<a$

$v=k x^3 \quad k=\frac{1}{2} m v^2$

$k x^3+\frac{1}{2} m v^2=k a^3$

$v= \frac{d x}{d t} =\sqrt{\frac{2 k}{m}\left(a^3-x^3\right)}$

$\int_0^{T / 2} d t =\sqrt{\frac{m}{2 k}} \int_0^a \frac{d x}{\sqrt{a^3-x^3}}$

$x =a \sin ^{2 / 3} \theta$

$d x =\frac{2}{3} a \sin ^{-1 / 3} \theta \cos \theta d$

$\frac{T}{2}=\sqrt{\frac{m}{2 k}} \frac{2}{3} \frac{1}{\sqrt{a}} \int_0^{\pi / 2} \sin ^{-1 / 3} \theta d \theta$

$T=\left[\frac{1}{3} \sqrt{\frac{m}{2 k}} \int_0^{\pi / 2} \sin ^{-1 / 3} \theta d \theta\right] \frac{1}{\sqrt{a}}$

$=\frac{1}{3} \sqrt{\frac{2 m}{k}} \frac{1}{\sqrt{a}}$

$\int_0^{\pi / 2} \sin ^{-1 / 3} \theta d \theta \approx 2.1$

${\left[T\right] \equiv[m]^\alpha[k]^\beta[A]^\gamma}$

$U=k|x|^3$

$M^{1} L^2 T^{-2}=k L^3$

$k=M^1 L^{-1} T^{-2}$

$\left[M^0 L^0 T^1\right]=\left[M^1\right]^\alpha\left[M^1 L^{-1} T^{-2}\right]^\beta[L]^\gamma$

$\alpha+\beta=0$, $-\beta+\tau=0$ and $-2 \beta=1$

$\beta=-\frac{1}{2}$, $\tau=-\frac{1}{2}$, and $\alpha=\frac{1}{2}$

$T \propto a^{-1 / 2}$