Embibe Experts Solutions for Chapter: Simple Harmonic Motion, Exercise 2: Exercise-2

A block of mass $m$ is resting on a piston which is moving vertically with a SHM of period $1 \mathrm{s}$. The minimum amplitude of motion at which the block and piston separate is: (Take $g=10 \mathrm{m} {\mathrm{s}}^{-2}$)

A point particle of mass $0.1 \mathrm{kg}$ is executing S.H.M. of amplitude of $0.1 \mathrm{m}.$ When the particle passes through the mean position, its kinetic energy is $8\times {10}^{-3} \mathrm{J}.$ The equation of motion of this particle when the initial phase of oscillation is $45°$ can be given by

Acceleration $a$ versus time $t$ graph of a body in SHM is given by a curve shown below. $T$ is the time period. Then corresponding graph between kinetic energy $\mathrm{KE}$ and time $t$ is correctly represented by,

On a smooth inclined plane, a body of mass $M$ is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has force constant $K$, the period of oscillation of the body (assuming the springs as massless) is

The displacement of a linear harmonic oscillator is given by $x=A \cos \omega t.$ The curves showing the variation of the potential energy with $t$ and $x$ (see figure) are displayed respectively by:

The mass $M$ shown in the figure oscillates in simple harmonic motion with amplitude $A.$ The amplitude of the point $P$ is 

Assertion: In a compound pendulum, if the suspension point and centre of oscillation are mutually interchanged, then no change in a time period is obtained.

Reason: Length of the equivalent simple pendulum remains same in both the case.

Assertion: In SHM, the motion is to and fro and periodic.

Reason: Velocity of the particle, $v=\omega \sqrt{a^2-x^2}$ (where $x$ is displacement)