Home
Class 12
PHYSICS
Two particles execute simple harmonic mo...

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. Their phase difference is

A

`pi/6`

B

0

C

`2pi/3`

D

`pi`

Text Solution

Verified by Experts

The correct Answer is:
C


Time interval `=T/6+T/6=(2T)/(6)`
Phase difference `implies (2T)/(6)-=(2pi)/(3)`
Promotional Banner

Similar Questions

Explore conceptually related problems

Two particles execute SHM of same amplitude and frequency on parallel lines. They pass one another when moving in opposite directions each time their displacement is half of their amplitude. What is the phase difference between them?

Two particles execute SHM of same amplitude and frequency on parallel lines. They pass one another when moving in opposite directions each time their displacement is one third their amplitude. What is the phase difference between them?

Two particles execute SHMs of the same amplitude and frequency along the same straight line. They cross one another when going in opposite direction. What is the phase difference between them when their displacements are half of their amplitudes ?

A particle executing simple harmonic motion with an amplitude A. The distance travelled by the particle in one time period is

Two particles are executing simple harmonic of the same amplitude (A) and frequency omega along the x-axis . Their mean position is separated by distance Xo. (Xo>A). If the maximum separation between them is (Xo+A), the phase difference between their motion is:

A particle executing simple harmonic motion with an amplitude A and angularr frequency omega . The ratio of maximum acceleration to the maximum velocity of the particle is

Two particles are in SHM with the same amplitude and frequency along the same line and about the same point. If the maximum separation between them is sqrt(3) times their amplitude, the phase difference between them is (2pi)/n . Find value of n

A particle is subjected to two simple harmonic motions of the same frequency and direction. The amplitude of the first motion is 4.0 cm and that of the second is 3.0 cm . Find the resultant amplitude if the phase difference between the two motion is (a) 0^(@) (b) 60^(@) ( c) 90 ^(@) (d) 180^(@)

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

A particle is executing simple harmonic motion with an amplitude A and time period T. The displacement of the particles after 2T period from its initial position is