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

`y(x,t)=0.06sin(2pi)/3**cos120pit`
(a) The displacement which involves harmonic function of `x` and `t` separately represents a stationary wave and the displacement, which is harmonic function of the form`(vt +_x)`,represents a travelling wave. Hence, the equation givenabove represents a stationary wave.
(b) When a wave pulse `y_(1)=-asin2pi/(lambda)(vt-x)` travelling along x-axis is super imposed by the reflected pulse.
`y_(2)=-asin2pi/lambda(vt+x)` from the other end. a stationary wave is formed and is given by
`y = y_(1)+y_(2)=-2asin 2pi/lambda**cos2pi/lambdavt`
Comparing the eqs.)i) and (ii). we have
`2pi/lambda=2pi/3 or lambda=3m`

`2pi/lambdav=120pi or v =60lambda=60**3=180ms^(-1)` Now frequency `gamma=v/lambda=180/3=60Hz`
(c) Velocity of transverse wave in astring is given by `v=sqrt(T/m)`
Here `v=180 m=(3**10^(-2))/1.5=2**10^(-2) kgm^(-1)` Also `v=180ms6(-1)`
`therefore T=v^(2)m = (180)^(2)**2**10^(-2)=648N`.