Simple Harmonic Motion

A spring of force constant $200 \mathrm{N} {\mathrm{m}}^{-1}$ is cut into three parts of lengths in ratio $1: 2: 3$. The smallest spring is further cut into two parts of equal length. The force constant of the shortest spring formed will be

At $t=\frac{2T}{3}$ a particle is at $\frac{-A}{2}$ and moving towards mean position. Find the equation of SHM. Also find the speed at $\frac{T}{4}$

Two particles executing SHM of same frequency and same amplitude meet at $x=+\frac{\sqrt{3}A}{2}$while moving in opposite directions. Phase difference between the particles is :-

A particle executes S.H.M. of amplitude $A$ along $x$-axis. At $t=0$, the position of the particle is $x=\frac{A}{2}$ and it moves along positive $x$-axis. The displacement of particle in time $t$ is $x=A \sin \left(\omega t + \delta \right)$, then the value $\delta$ will be

For a periodic motion represented by the equation $y=\mathrm{sin\omega }t+\mathrm{cos\omega }t$ the amplitude of the motion is

If initially a particle is at$x=\frac{-A}{\sqrt{2}}$ and its position is given by $x=A\cos \left(\omega t+\varphi \right)$ ) then find $\varphi$ if the particle is moving away from the mean position?

A particle moves on the$x$-axis according to the equation $x=\left(5 + 2 \sin \omega t\right) \mathrm{m}$.The motion is simple harmonic with amplitude

The displacement of a particle executing simple harmonic motion is given by$y=A_0+A\sin \omega t+B\cos \omega t$.Then the amplitude of its oscillation is given by

The shortest distance travelled by a particle executing SHM from extreme position in$3$seconds is equal to half of its amplitude. The time period of given particle is

Describe the motion corresponding to $x-t$equation, $x=10-4\cos \omega t$

Two simple harmonic motions are represented by equations $y_1=4\sin (10t+\varphi )$ and $y_2=5\cos 10t.$ What is the phase difference between their velocities?
 

Derive an expression for displacement of a particle performing linear simple harmonic motion.

A body of mass $1 \mathrm{kg}$ is suspended from massless spring of natural length $1 \mathrm{m}$. If mass is released from rest, spring can stretch up to a maximum vertical distance of $40 \mathrm{cm}$, the potential energy stored in the spring at this extension is ($\mathrm{g}=10 {\mathrm{ms}}^{-2}$)

A block of mass $0.1 \mathrm{kg}$ attached to a spring of spring constant $400 {\mathrm{Nm}}^{-1}$ is pulled horizontally from$x = 0$ to $x_1 = 10 \mathrm{mm}$. Find the work done by spring force.
 

The work done by the tension in the string of a simple pendulum in one complete oscillation is equal to

Displacement of a particle is given by $y=0.2\sin \left(10\mathrm{\pi }t+1.5\mathrm{\pi }\right)$. Is it simple harmonic? If so, what is its period?

Two linear simple harmonic motions with same amplitude and frequency $\omega$ and $2\omega$ get superimposed on a particle in $x$ and $y$ direction. If the phase difference between the two are $\frac{\pi }{2}$ initially, the resultant path will be:

Equation of Displacement in SHM
The displacement $y\left(t\right)=A\sin \left(\omega t+\varphi \right)$ of a pendulum for $\varphi =\frac{2\pi }{3}$ is correctly represented by

The displacement of a particle executing SHM is given by $X=3\sin \left[2\pi t+\frac{\pi }{4}\right]$ where $x$ is in meters and $t$ is in seconds. The amplitude and maximum speed of the particle is

An object undergoing simple harmonic motion takes $0.5 \mathrm{s}$ to travel from one point of zero velocity to the next such point. The angular frequency of the motion is,