A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidal. The amplitude of vibration is 1.0cm and the displacement becomes zero 200 times per second. The linear mass density of the string is `0.10 kg m^(-1)` and it is kept under a tension of 90 N. (a) Find the speed and the wavelength of the wave. (b) Assume that the wave moves in the positive x-direction and at t = 0 the end x= 0 is at its positive extreme position. Write the wave equation. (c ) Find the velocity and acceleration of the particle at x = 50 cm at time t = 10ms.

Answer

A, B, C, D

Explanation

Amplitude, A = 1 cm
Tension, T = 90 N
Frequency, `f = (200)/(2) = 100 Hz`
Mass per unit length
m = 0.1 kg//m
(a) `V = ((T)/(m))`
`=sqrt((90)/(0.1)) = 30m//s`
`lambda = (V)/(f) = (30)/(100)`
`=0.3 = 30 cm
(b) The wave equation,
`y = (1cm) cos 2pi {((t)/(0.01s)) - ((x)/(30)cm)}`
`[because at x = 0 displacement is maximum]
(c ) `y = 1cos 2pi ((x)/(30) - (t)/(0.01))`
`rArr upsilon = (dy)/(dt)`
`=((1)/(0.01))2pi cos 2pi {((x)/(30)) - ((t)/(0.01))}`
`a = (dupsilon)/(dt)`
` = {(4pi^2)/((0.01)^2)}sin 2pi {((x)/(30)) - ((t)/(0.01))}`
When, x = 50 cm, t = 10 ms = 10 xx10xx10^(-3)s`
`x = ((2pi)/(0.01))sin 2pi{((5)/(3)) - ((0.01)/(0.01))}`
`= - ((pi)/(0.01))sin ((2pixx (2)/(3))`
`=-((1)/(0.01))sin ((4pi)/(3))`
`=200 pi sin ((pi)/(3))`
`200 pi xx (sqrt3)/(2)`
`=544 cm//s = 5.4 m//s`
Similarly,
`a = {(4pi^2)/((0.01^2)} cos 2pi {((5)/(3)) - 1}`
`=4pi^2 xx 10^4 xx (1)/(2)`
`=2xx 10^5 cm//s^2 = 2km//s^2`