Home
Class 12
PHYSICS
A particle performing SHM starts equi...

A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is `pi m//s`. Amplitude of oscillation is
`(cos 45^(@) = 1/(sqrt(2)))`

A

`2sqrt2`m

B

`4sqrt2`m

C

`6sqrt2`m

D

`8sqrt2`m

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Similar Questions

Explore conceptually related problems

A body starting from mean position is executing simple harmonic motion. Its time . period is 24 s. After 4 s, its velocity is pi" "m//s then its path length is

Particle performing S.H.M. starts from mean position. Plot a graph of displacement, velocity and acceleration.

A particle is in S.H.M. along a straight line 0.2 m long with a period of 6 s. Its displacement after 1/2 second, if its epoch is ((pi)/(6))^c , will be

A particle executing S.H.M. starts from the mean position. Its phase, when it reaches the extreme position, is

The phase of a particle performing S.H.M. increases by pi//2 after every 4 seconds. Its time period of oscillation is

A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and time period is T. At the instance when its speed is half the maximum speed , its displacement x is

A particle performing linear S.H.M has maximum velocity of 25 cm/s and maximum acceleration of 100 cm/s^2 . Find the amplitude and period of oscillation . (pi = 3.142)

A particle performs SHM of amplitude 10 cm. Its maximum velocity during oscillation is 100 cm/s. What is its displacement, when the velocity is 60 cm/s.

If a simple pendulum oscillates with an amplitude of 50 mm and time period of 2 sec, then its maximum velocity is

If the particle in linear S.H.M. starts from the extreme left position, then its equation of motion is given by