Question

A transverse wave is described by the equation Y=Y0 sin 2(pie)(ft - x/ lamda). Prove that the maximum particle velocity is equal to four times the wave velocity if
lamda=(pie)yn/2

Answer 1

given,

Y = Y° sin2π(ft -x/lemda )
particle velocity is dY/dt

so, differentiate equation wrt t

dY/dt = 2πY°f cos2π(ft -x/lemda )
maximum velocity of particle =2πY°f

given, lemda = πY°/2

but we know ,
velocity of wave = frequency × wavelength = π/2Y°f

now ,
velocity of particle/velocity of wave = 2πY°f/πY°f/2 =4

hence,
max velocity of particle = 4 × velocity of wave

Answer 2

see attachment....I hope it will help u...