Oscillations Te...

A body executing SHM at a displacement 'x' its PE is $$E_1$$, at a displacement 'Y' its PE is $$E_2$$ The P.E at a displacement $$(x+y)$$ is  

The vertical extension in a light spring by a weight of $$1\ kg$$ suspended from the wire is $$9.8\ cm$$. The period of oscillation: 

A particle execute SHM with time period $$T$$ and amplitude $$A$$. The maximum possible average velocity in time $$\dfrac {T}{4}$$:

If y denotes the displacement and t denote the time and the displacement is given by $$y=a\,sin\,\omega t$$, the velocity of the particle is 

The equation of S.H.M of a particle of amplitude 4 cm performing 150 oscillations per minute starting with an initial phase $$30^0$$ is 

A particle execute S.H.M from the mean position. its amplitude is $$A$$, its time period is "T'. At what displacement,its speed is half of its maximum speed.

The displacement of a particle in S.H.M. is indicated by equation $$y\ = \ 10\ sin(20t\ +\ \pi/3)$$ where $$y$$ is in metres. The value of time period of vibration will be (in seconds):

A particle is oscillating according to the equation $$X = 7\cos 0.5\pi t$$, where $$t$$ is in seconds. The point moves from the position of equilibrium to maximum displacement in time.

The maximum velocity of a particle performing linear S.H.M. is $$0.16\ m/s$$ and its maximum acceleration is $$0.64\ m/s^{2}$$. The calculate its time period

Two particles are executing simple harmonic motion with same angular frequency but different amplitudes $$A$$ and $$2A$$ about same mean position on the same straight line. They cross each other when they are at a displacement of $$\dfrac{4A}{5}$$ from mean position in same direction. Then phase difference between them is:-