A transverse wave is travelling on a string and is governed by the equation `y= 2 sin(omegat – kx)`. The wavelength for which wave velocity will be equal to maximum particle velocity is _____m.

Answer

12.56

Explanation

Wave velocity, `v = omega/k`
Particle velocity, `v_p=(dy)/(dt)=Aomegacos (omegat-kx)`
`:. (V_p)_(max) = Aomega`
`Aomega=omega/k`.....(given)
`implies A = lamda/(2pi)`
`implies lamda = 2piA = 4pi = 12.56 m`