Two particle execute `SHM` with amplitude `A` and `2A` and angular frequency `omega` and `2omega` repectively. At `t=0` they starts with some initial phase difference. At, `t=(2pi)/(3omega)`. They are in same phase. Their initial phase difference is-

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 particle starts from rest at the extreme position and makes SHM with an amplitude A and angular frequency omega . Its displacement at time 't' is

Two linear SHM of equal amplitudes A and frequencies omega and 2omega are impressed on a particle along x and y - axes respectively. If the initial phase difference between them is pi//2 . Find the resultant path followed by the particle.

Two linear SHM of equal amplitudes A and frequencies omega and 2omega are impressed on a particle along x and y - axes respectively. If the initial phase difference between them is pi//2 . Find the resultant path followed by the particle.

y_1 =4sin( omega t + kx), y2 = -4 cos( omega t + kx), the phase difference 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: