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

A simple pendulum with length $L$ and bob of mass $m$ executes SHM of small amplitude $A$. The maximum tension in the string will be 

A $25 \mathrm{kg}$ uniform solid sphere with a $20 \mathrm{cm}$ radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. A torque of $0.10 \mathrm{N} \mathrm{m}$ is required to rotate the sphere through an angle of $1.0 \mathrm{rad}$and then the orientation is maintained. If the sphere is then released, its time period of the oscillation will be:

A metre stick swinging about its one end oscillates with frequency $f_0$. If the bottom half of the stick was cut off, then its new oscillation frequency will be:

When two mutually perpendicular simple harmonic motions of same frequency, amplitude and phase are superimposed 

The position of a particle in motion is given by $y=C\sin \omega t+D\cos \omega t$ with respect to origin. Then motion of the particle is:

A simple harmonic motion is given by $y=2(4\sin 3\pi t+3\cos 3\pi t).$ What is the amplitude of motion if $y$ is in $m$?

The position vector of a particle moving in $x-y$ plane is given by $\overrightarrow{r}=\left(A\sin \omega t\right)\hat{\mathrm{i}}+\left(A\mathrm{cos\omega }t\right)\hat{\mathrm{j}}$, then motion of the particle is:

When a damped harmonic oscillator completes $100$ oscillations, its amplitude is reduced to ${\left(\frac{1}{3}\right)}^{\mathrm{rd}}$ of its initial value. What will be its amplitude when it completes $200$ oscillations?