Embibe Experts Solutions for Chapter: Simple Harmonic Motion, Exercise 1: Exercise-1

Equation of SHM is $x=10\sin 10\pi t$. Find the distance between the two points where speed is $50\pi \mathrm{cm} {\sec }^{-1}$. $x$ is in $\mathrm{cm}$ and $t$ is in seconds.

A mass $M$ is performing linear simple harmonic motion, then correct graph for acceleration $a$ and corresponding linear velocity $v$ is

The amplitude of a particle executing SHM with the frequency of $60 \mathrm{Hz}$ is $0.01 \mathrm{m}$. The maximum value of the acceleration of the particle is

A particle is executing SHM of frequency $300 \mathrm{Hz}$ and with amplitude $0.1 \mathrm{cm}$. Its maximum velocity will be

The velocity of a particle in SHM at displacement $y$ from the mean position is ($a=$amplitude, $\omega =$angular frequency).

The $x-t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t=\frac{4}{3} \mathrm{s}$ is

The amplitude of a particle executing SHM with the frequency of $60 \mathrm{Hz}$ is $0.01 \mathrm{m}$. The maximum value of the acceleration of the particle is

The displacement of an oscillating particle varies with time (in $\mathrm{seconds}$) according to the equation, $y \left(\mathrm{cm}\right)=\sin \frac{\mathrm{\pi }}{2}\left(\frac{t}{2}+\frac{1}{3}\right)$. The maximum acceleration of the particle is approximately