Transverse waves on a srting have speed `12.0 m//s`, amplitude `0.05 m` and wavelength `0.4 m`. The waves travel in the `+ x-direction` and at `t = 0` the `x = 0` end of the string has zero displacement and is moving upwards.
(a) Write a wave function describing the wave.
(b) Find the transverse displacement of a point at `x = 0.25 m` at time `t = 0.15 s`.
( c ) How much time must elapse from the instant in part (b) until the point at `x = 0.25 m` has zero dispacement?

Answer

A, B, C, D

Explanation

(a) `omega = 2pif = 2pi((v)/(lambda)) = 2pi((12)/(0.4))`
`=(60 pi) rad//s`
`k = (2pi)/(lambda) = (2pi)/(0.4) = (5pi) m^(-1)`
Since, wave is travelling along `+ve` x-direction, `omegat and kx` should have opposite sign. Further at `t = 0 , x = 0` the string has zero displacement and moving upward (in positive cirection). Hence at `x = 0`, we should have `Asin omegat` not `- Asin omegat`. Therefore, the correct expression is
`y A sin (omegat - kx)`
or `y = 0.05 sin (60pit - 5pix)`
(b) Putting `x = 0.2 m` and
`t = 0.15 s` in above equation we have, `y = -0.035 4m`
`= -3.54 cm`
( c ) In part (b), `y =A// sqrt(2)`
From `y =(A)/(sqrt(2))` to `y = 0`, time taken is
`t = (T)/(8) = (2pi)/(omega xx 8)`
`=(2pi)/((60pi)(8))`
`= 4.2 xx 10^(-3) s`
`=4.2 ms`