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 given eqn. is `y(x,t)=0.06sin((2pi)/(3))xcos120pit` ...(i)
(i) A s the equation involves harmonic functions of x and t separately, it represents a stationary wave.
(ii) We know that when a wave pulse `lambda_(1)=rsin((2pi)/(lambda))(upsilont-x)`
travelling along `+` direction of x-axis is superimposed by the reflected wave
`y_(2)=-rsin((2pi)/(lambda))(upsilont-x)`
travelling in opposite direction, a stationary wave
`y=y_(1)+y_(2)=-2sin((2pi)/(lambda))x cos((2pi)/(lambda))upsilont ` is formed ...(ii)
Comparing (i) and (ii), we fing that `(2pi)/(lambda)=(2pi)/(3)o rlambda=3m`
Also, `(2pi)/(lambda)upsilon=120pi or upsilon=60lambda=60xx3=180ms^(-1)`
Frequency, `v=(upsilon)/(lambda)=(180)/(3)=60hertz`
Note that both the waves have same wave length, same frequency and same speed.
(c) Velocity of transverse wave is `upsilon=sqrt(T//m) or upsilon^(2)=T//m`
`T=upsilon^(2)xxm, ` where ` m =(3xx10^(-2))/(1.5)=2xx10^(-2)kg//m`
`:. T=(180)^(2)xx2xx10^(-2)=648N`