Home
Class 12
PHYSICS
Two particle undergoes SHM along paralle...

Two particle undergoes `SHM` along parallel line with the same time period `(T)` and equal amplitude At a particular instant , one particle is at its extereme position while the other is at its mean position .They move in the same direction .They will cross each other after a further time

A

T/8

B

3T/8

C

T/6

D

4T/3

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • PRACTICE WORKSHEET 1

    D MUKHERJEE|Exercise Assertion- Reason Type|3 Videos
  • PRACTICE WORKSHEET 1

    D MUKHERJEE|Exercise Linked- Comprehension Type|3 Videos
  • OPTICS

    D MUKHERJEE|Exercise All Questions|47 Videos
  • PRACTICE WORKSHEET 2

    D MUKHERJEE|Exercise Linked- Comprehension Type|4 Videos

Similar Questions

Explore conceptually related problems

Two particles undergo SHM along the same line with the same time period (T) and equal amplitude (A). At a particular instant one is at x = -A and the other is at x = 0. If they start moving in the same direction then they will cross each other at (i) t = (4T)/(3) (ii) t = (3T)/(8) (iii) x = (A)/(2) (iv) x = (A)/(sqrt(2))

Two particles undergo SHM along the same line with the same time period (T) and equal amplitude (A) . At a particular instant one is at x =- A and the other is at x =0 . If they start moving in the same direction then they will cross each other at (i) t = (4T)/(3) (ii) t = (3T)/(8) (iii) x = (A)/(2) (iv) x = (A)/(sqrt(2))

Time period (T) and amplitude (A) are same for two particle which undergoes SHM along the same line. At one particular instant one particle is at phase (3pi)/(2) and the other is at zero,while moving in the same direction. Find the time at which they will cross each other

At a particular position the velocity of a particle in SHM with amplitude a is sqrt3 // 2 that at its mean position. In this position, its displacement is :

Two very small particles are executing SHM along very near parallel lines about the same mean position, same amplitude and the same frequency. At t = 0 both the particle are moving with speed of 1m//sec but on the opposite side of the mean position and having same direction of velocity at a distance of 5cm from the mean position. At t = 4sec they are at the same position moving with speed of 1m//sec but in opposite direction. Assuming that this is the minimum time form t = 0 where they are meeting for the 1st time. If the maximum speed of either particle is (xpi)/(4sqrt(2))cm//sec Find x

A particle performs SHM on a straight line with time period T and amplitude A . The average speed of the particle between two successive instants, when potential energy and kinetic energy become same is

A particle performs SHM with a period T and amplitude a. The mean velocity of particle over the time interval during which it travels a//2 from the extreme position is