If the maximum speed of a particle on a travelling wave is `v_(0)`, then find the speed of a particle when the displacement is half of the maximum value.

Options

A. `(v_(0))/(2)`
B. `sqrt(3v_(0))/(4)`
C. `sqrt(3 v_(0))/(2)`
D. `v_(0)`

Answer

c

Explanation

For the wave `y=A sin(omegat-kx),v_(0)=Aomega`
where A is, the maximum displacement.for the given condition,
`(A)/(2)=A sin(omegat-kx)`
`sin(omegat-kx)=(1)/(2)`
And `(dely)/(delt)=Aomega cos (omegat-kx)=Aomegasqrt(3)/(2)=sqrt(3v_(0))/(2)`