A transverse sinusoidal wave of amplitude `a`, wavelength `lambda` and frequency `f` is travelling on a stretched string. The maximum speed of any point in the string is `v//10`, where `v` is the speed of propagation of the wave. If `a = 10^(-3)m` and `v = 10ms^(-1)`, then `lambda` and `f` are given by

Options

A. `lamda = 2 pi xx 10^-2 m`
B. `lamda = 10^-3 m`
C. `f = (10^3)/(2 pi) Hz`
D. `f = 10^4 Hz`

Answer

A, C

Explanation

For a transverse sinusoidal wave travelling on a string the maximum velocity is `a omega`. Also, the maximum velocity is
`(v)/(10)=(10)/(10) =1m//s`
`:. a omega =1 rArr 10^(-3) xx 2 pi f=1`
`rArr f =(1)/(2pi xx 10^(-3)) =(10^(3))/(2 pi) Hz`
The velocity `v = f lamda`
`:. lamda = (v)/(f) =(10)/(10^(3)//2pi) = 2pi xx 10^(-2) m`.