Home
Class 12
PHYSICS
A point performs simple harmonic oscilla...

A point performs simple harmonic oscillation of period T and the equation of motion is given by `x = a sin (omega t + (pi)/(6))`. After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity ?

A

`T/3`

B

`T/12`

C

`T/8`

D

`T/6`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Similar Questions

Explore conceptually related problems

What is the time - period of x = A sin (omega t + alpha) ?

Two simple harmonic motions given by, x = a sin (omega t+delta) and y = a sin (omega t + delta + (pi)/(2)) act on a particle will be

A simple harmonic oscillation has an amplitude A and time period T . The time required to travel from x = A to x= (A)/(2) is

A simple harmonic motino has amplitude A and time period T. The maxmum velocity will be

A particle performs simple harmonic motion about O with amolitude A and time period T . The magnitude of its acceleration at t=(T)/(8) s after the particle reaches the extreme position would be

A body is performing simple harmonic motion with amplitude a and time period T variation of its acceleration (f) with time (t) is shown in figure If at time t velocity of the body is v which of the following graph is correct?

A body is performing simple harmonic motion with amplitude a and time period T variation of its acceleration (f) with time (t) is shown in figure If at time t velocity of the body is v which of the following graph is correct?

The displacement of a particle executing simple harmonic motion is given by y = 4 sin(2t + phi) . The period of oscillation is

A particle executes a simple harmonic motion of time period T. Find the time taken by the particle to go directly from its mean position to half the amplitude.

The motion of a particle executing simple harmonic motion is given by X = 0.01 sin 100 pi (t + 0.05) , where X is in metres andt in second. The time period is second is