The transverse displacement of a string (clamped at its two ends ) is given by
`y(x,t)=0.06sin((2pi)/(3))xcos(120pit)`
wherer x ,y are in m and t ini s. The length of the string is 1.5m and its mass is `3xx10^(-2)` kg. Answer the following: (i) Does the function represent a travelling or a stationary wave ?
(ii) Interpret the wave as a superimposition of two waves travelling in opposite directions. What are the wavelength, frequency and speed of propagation of each wave ?
(iii) Determing the tension in the string.

Explanation

The transverse displacement of a string is given by
`y(x, t) = 0.060 "sin" (2pi)/(3) x cos (120 pi t)`
(a) The given equation is similar to the wave equation `y (r, t) = 2 a sin kx cos omega t`
As the equation involves harmonic functions of x and t, therefore, it represents a stationary wave.
(b) `y(x, t) = "0.060 sin" (2pi)/(3) x cos (120 pi t)" "...(i)`
When a wave pulse `y_(1) = "r sin" (2pi)/(lambda) (upsilon t - x)` travelling along positive direction of X-axis is superimposed by the reflected wave
`y_(2) = "r sin" (2pi)/(lambda) [vt + x]`
travelling in opposite direction then the resultant wave is a stationary wave as
`y = y_(1) + y_(2)`
`= - "2 r sin" (2pi)/(lambda) "x cos" (2pi)/(lambda) vt`
Comparing with equation (i)
`(2pi)/(lambda) = (2 pi)/(3)` therefore, `lambda = 3` m
`(2pi)/(lambda) upsilon = 120 pi` therefore, `upsilon = 180` m/s
Hence, frequency `v = (upsilon)/(lambda) = (180)/(3) = 60 s^(-1)`
(c) `v = (1)/(2l) sqrt((T)/(m))`
Here `m = (3.0 xx 10^(-2) kg)/(1.5 m) = 2 xx 10^(-2) kg m^(-1)`
`60 = (1)/(2 xx 1.5) sqrt((T)/(2 xx 10^(-2)))`
Squaring both sides,
`60 xx 60 = (1)/(4 xx 1.5 xx 1.5) xx (T)/(2 xx 10^(-2))`
Solving, we get T = 648 N.