The transverse displacement of a string ( clamped at its both ends ) is givenby `y ( x,t) = 0.06 sin((2pi)/(3) x) cos ( 120 pi t )`
where x and y are in m and t in s. The length of the string is 1.5m and its mass is `3.0 xx 10^(-2) kg `
Answer the following `:`
(a) Does the function represent a travelling wave or a stationary wave ?
(b) Interpret the wave as a superposition of two waves travelling in opposite directios. What is the wavelength, frequency,and speed of each wave ?
(c ) Determine the tension in the string.

Explanation

The given equation is
`y( x,t )=0.06 sin .( 2pi)/( 3) xcos 120 pit `
(a) As the equaton involves harmonic functions of xand t separatel, it represents a stationary wave.
(b) We know that when a wave pulse `y_(1)= r sin. ( 2pi)/(lambda) (vt +x) ` ...(i)
travelling along `+` direction of x-axis is super improsed by the reflected wave
`y = y_(1)+y_(2) = - 2 rsin. ( 2pi)/( lambda) x cos. ( 2pi)/( lambda) vt ` is formed.
Comparing (i) & (ii) wefindthat
`(2pi)/( lambda) = ( 2pi)/( 3) implies lambda= 3m`.
Also`( 2pi)/(lambda) v= 120 pi ` (Or)
`V = 60lambda =60 xx 3 = 180 m//s `
frequency , `v = ( v )/( lambda) = (180)/( 3) = 60` hertz
Note that both the wave length, same frequency and same speed.
(c ) Velocity of transverse waves is
`v = sqrt((T)/( m)) ` (or) `v^(2) = T//mu`
`T = V^(2) xxmu ` where `mu= ( 3xx 10^(-2))/( 1.5 ) `
`=2 xx10^(-2) kg//m`
`T=( 180)^(@) xx 2 xx 10^(-2) = 648N`