Home
Class 12
PHYSICS
Two particle are in SHM on same striaght...

Two particle are in `SHM` on same striaght line with amplitude `A` and `2A` and with same angular frequency `omega`. It is observed that when first particle is at as distance `(A)/sqrt(2)` from origin and going toward mean position, other particle is at extreme postion on other side of mean positon. find phase difference betwenen he two paricles.

A

`45^@`

B

`90^@`

C

`135^@`

D

`180^@`

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    MOTION|Exercise EXERCISE -1 (Objective Problems | NEET) SECTION-D Spring System and Spring Pendulum|4 Videos
  • SIMPLE HARMONIC MOTION

    MOTION|Exercise EXERCISE -1 (Objective Problems | NEET) SECTION-E Simple Pendulum|6 Videos
  • SIMPLE HARMONIC MOTION

    MOTION|Exercise EXERCISE -1 (Objective Problems | NEET) SECTION-B Displacement, Velocity and Acceleration in S.H.M|26 Videos
  • SEMI CONDUCTOR AND LOGIC GATES

    MOTION|Exercise EXERCISE 3|34 Videos
  • SIMPLE HARONIC MOTION

    MOTION|Exercise EXERCISE -4 (Leve-II) [ JEE 2011]|11 Videos

Similar Questions

Explore conceptually related problems

Write equation of SHM of angular frequency omega and amplitude x_(m) if the particle is situated at x_(m)//sqrt2 at t= 0 and is going toward mean position.

Two particles are in SHM with same amplitude A and same regualr frequency omega . At time t=0, one is at x = +A/2 and the other is at x=-A/2 . Both are moving in the same direction.

Two particles move parallel to x- axis about the origin with same amplitude 'a' and frequency omega . At a certain instant they are found at a distance a//3 from the origin on opposite sides but their velocities are in the same direction. What is the phase difference between the two?

Two particles P and Q are executing SHM across same straight line whose equations are given as y_(P)=Asin (omegat+phi_(1)) and y_(Q)=Acos (omegat+phi_(2)) An observer, at t=0 observer the particle P at a distance A//sqrt(2) moving to the right from mean position O while particle Q at (sqrt(3))/(2) A moving to the left from mean position O as shown in figure. Then phi_(2)-phi_(1)(i n rad) ie equal to

In previous question, if the second pendulum bob is thrown at velocity v_(2) at an angle beta from mean position, but on other side of mean position , find the phase diffrence in the two SHMs now as show in the figure.

Two particles execute SHM of the same time period along the same straight lines. They cross each other at the mean position while going in opposite directions. Their phase difference is

Two particles are in SHM along same line with same amplitude A and same time period T. At time t=0, particle 1 is at + A/2 and moving towards positive x-axis. At the same time particle 2 is at -A/2 and moving towards negative x-axis. Find the timewhen they will collide.

Two particles executing SHM of same frequency, meet at x= +A//2 , while moving in opposite direction . Phase difference between the particles is

At particle is executing SHM with amplitude A and has maximum velocity v_(0) . Find its speed when it is located at distance of (A)/(2) from mean position.

Two particles are executing SHM in a straight line. Amplitude A and the time period T of both the particles are equal. At time t=0, one particle is at displacement x_(1)=+A and the other x_(2)=(-A/2) and they are approaching towards each other. After what time they across each other? T/4