The eqyation of wave is ` x= 5sin ((t)/( 0.4) -( x)/(4))cm ` the maximum velocity of the particles of the medium is

Options

A. 1 m/s
B. 1.5 m/s
C. 1.25 m/s
D. 2 m/s

Answer

C

Explanation

The given equation of wave is
`y=5sin((t)/(0.04)-(x)/(4))cm`
The particle velocity of wave is
`v_(P)=(dy)/(dt)=5((1)/(0.04))cos((t)/(0.04)-(x)/(4))cm//s`
`costheta` has maximum value `+1`.
For maximum particle velocity `cos((t)/(0.04)-(x)/(4))=+1`,
`therefore (v_(P))_(max)=(5)/(0.04)cm//s=125cm//s=1.25cm//s`