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)=0.06"sin"(2pix)/(3)xxcos120pit` . . (i)
(i) The displacement which involves harmonic functions of x and t separately represents a stationary wave and the displacement, which is harmonic function of the form `(vt+-x)`, represent a travelling wave. hence, the equation given above represents a stationary wave.
(ii) When a wave pulse `y_(1)=a"sin"(2pi)/(lamda)(vt-x)` travelling along x-axis is superimposed by the reflected pulse.
`y_(2)=-"sin"(2pi)/(lamda(vt+x)` from the other end, a stationary wave is formed and is given by
`y=y_(1)+y_(2)=-2a" sin"(2pi)/(lamda)xx"cos"(2pi)/(lamda)vt` . . . (ii)
Comparing the eqs. (i) and (ii) we have
`(2pi)/(lamda)=(2pi)/(3) or lamda=3m`
`(2pi)/(lamda)v=120pi or v=60lamda=60xx3=180ms^(-1)`
Now frequency `v=(v)/(lamda)=(180)/93)=60Hz`
(iii) Velocity of transverse wave in a string is given by `v=sqrt((T)/(mu))`
Here, `mu=(3xx10^(-2))/(1.5)=2xx10^(-2)kgm^(-1)`
Also, `v=180ms^(-1)" "therefore T=v^(2)mu=(180)^(2)xx2x10^(-2)=648N`