The transvers displacement of a string (clamped at its both ends) is given by
`y(x,t) = 0.06 sin ((2pi)/(3)s) cos (120 pit)`
Where `x` and `y` are in `m` and `t` in `s`. The length of the string `1.5 m` and its mass is `3.0 xx 10^(-2) kg`.
Answer the following :
(a) Does the funcation represent a travelling wave or a stational wave ?
(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength. Frequency and speed of each wave ?
Datermine the tension in the string.

Explanation

(a) The general equation representing a stationary wave is given by the displacement function:
`y(x,t)=2a"sin"kx "cos" omegat`
This equation is similar to the given equation
`y(x,t)=0.06 "sin"((2)/(3)x) "cos"(120 pit)`
Hence , the given function represents a stationary wave.
(b) A wave travelling along the positive x-direction is given as:
`y_(2)=a"sin"(omegat+kx)`
The superposition of these two waves yields:
`y=y_(1)+ y_(2) = a"sin"(omegat-kx)-a"sin"(kx) "cos"(omegat)-a"sin "(kx)" cos"(omegat)`
`=-2a"sin"(kx)"cos"(omegat)`
`=-2a"sin"(kx)"cos"(omegat)`
`=-2a"sin"((2pi)/(lambda)x)"cos"(2pivt)" "...(i)`
The transverse displacement of the string is given as:
`y(x,t)=0.06"sin"((2pi)/(3)x)"cos"(120 pit)" "...(ii)`
Comparing equations (i) and (ii), we have:
`(2pi)/(lambda)=(2pi)/(3)`
`therefore` Wavelength, `lambda=3m`
It is given that
`120pi=2 piv`
Frequency, ν = 60 Hz
Wave speed, v =`vlambda`
`=60xx3 = 180 m//s`
(c) The velocity of a transverse wave travelling in a string is given by the relation:
`v=sqrt((T)/(mu))" "...(i)`
Where,
Velocity of the transverse wave, v = 180 m/s
Mass of the string, `m = 3.0 xx 10^(-2)` kg
Length of the string, l = 1.5 m
Mass per unit length of the string, `mu=(m)/(l)`
`=(3.0)/(1.5)xx10^(-2)`
`=2xx10^(-2) kg m^(-1)`
Tension in the string = T
From equation (i), tension can be obtained as:
`T=v^(2)mu`
`=(180)^(2)xx2xx10^(-2)`
`=648 N`