Home
Class 11
PHYSICS
Two particle are executing simple harmon...

Two particle are executing simple harmonic motion. At an instant of time t their displacement are `y_(1)=a "cos"(omegat)` and `y_(2)=a "sin" (omegat)`
Then the phase difference between `y_(1)` and `y_(2)` is

Promotional Banner

Similar Questions

Explore conceptually related problems

The two waves represented by y_1=a sin (omegat) and y_2=b cos (omegat) have a phase difference of

Two waves are given by y_(1)=asin(omegat-kx) and y_(2)=a cos(omegat-kx) . The phase difference between the two waves is -

Two waves are given by y_(1)=asin(omegat-kx) and y_(2)=a cos(omegat-kx) . The phase difference between the two waves is -

Two waves are given by y_(1)=asin(omegat-kx) and y_(2)=a cos(omegat-kx) . The phase difference between the two waves is -

Two waves are represented by Y_1=a_1 cos (omegat-kx) and Y-2 =a_2 sin (omegat-kx+pi//3) Then the phase difference between them is-

Two particle A and B execute simple harmonic motion according to the equation y_(1) = 3 sin omega t and y_(2) = 4 sin [omega t + (pi//2)] + 3 sin omega t . Find the phase difference between them.

Two partical A and B execute simple harmonic motion according to the equation y_(1) = 3 sin omega t and y_(2) = 4 sin [omega t + (pi//2)] + 3 sin omega t . Find the phase difference between them.

Two SHMs are respectively represented by y_(1)=a sin (omegat-kx) and y_(2)=b cos(omegat-kx) . The phase difference between the two is

Two simple harmonic motions are represented by y_(1)= 10 "sin" omega t " and " y_(2) =15 "cos" omega t . The phase difference between them is