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

Options

A. `lamda=2pixx10^(-2)m`
B. `lamda=10^(-3)m`
C. `f=(10^(3))/(2pi)Hz`
D. `f=10^(4)Hz`

Answer

A, C

Explanation

For a transverse sinusoidal wave traveling on a string, the maximum velocity is `aomega` Also,the maximum velocity is
`v/10=10/10=1m//s therefore a omega=1implies 10^(-3)xx2pif=1 implies f=1/(2pixx10^(-3))=10^(3)/(2pi) Hz`
The velocity `v=flambda therefore lambda=v/f=(10)/(10^(3)//2pi)=2pixx10^(-2)m`