A transverse harmonic disturbance is produced in a string. The maximum transverse velocity is `3m//s` and maximum transverse acceleration is `9om//s`. If the wave velocity is `20 m//s` then find the waveform.

Answer

A, B, C

Explanation

KEY CONCEPT : The wave form of a transverse harmonic disturbance is
`y = a sin(omegat +- kx +- Phi)`
Given `v_(max) = aomega = 3m//s` …(i)
`A_(max) = aomega^(2) = 90m//s` …(ii) Velocity of wave `v = 20m//s` …(iii)
Dividing (ii) by (i)
`(aomega^(2))/(aomega) = (90)/(3)` rArr `omega = 30rad//s` ...(iv)
Substituting the value of `omega` in (i), we get
`a = (3)/(30) = 0.1m` ...(v)
Now, `k = (2pi)/(lambda) = (2pi)/(v//v) = (omega)/(v) = (30)/(20) = (3)/(2)` ...(vi)
From (iv), (v) and (vi) the wave form is
`y = 0.1 sin [30t +- (3)/(2) x +- Phi]`