The transverse displacement of a string (clamped at its both ends) is given by
`y(x, t) = 0.06 sin ((2 x)/(3) x) cos (120 pi t)`
where x and y are in m and t in s. The length of the string is 1.5 m 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 directions. What is the wavelength, frequency , and speed of each wave ?
(c ) Determine the tension in the string.

Explanation

(a) For of given wave equatino is not like `y =a sin (omega t pm kx + phi).` Hence, it does not represent travelling wave.
`y =0.06 sin ((2pi)/(3)x) cos (120 pi t)" "…(1)`
represents a stationary wave in which there is a product of two harmonic functions one in terms of positon x and another in terms of time t.
(b) Comparing equation (1) with stndard stationary wave equation,
`y = 2a sin (kx) cos (omega t ) " "...(2)`
we get `2a = 0.06 impliesa ==0.03 m " "...(3)`
`k = (2pi)/(lamda) = (2pi)/(3) implies lamda = 3m " "...(4)`
`omega =23pi f = 120 pi implies f = 60 Hz " "...(5)`
Equation of two component waves in a given statonary wave are :
`y _(1) =a sin (omega t -kx) =0.03 sin (120 pi t -(2pi)/(3) x)" "...(6)`
`y _(2) =- a sin (omega t + kx) =-0.03 sin (120 pi t + (2pi)/(3) x)" "...(7)`
Now, speed of propagation of wave is,
`v = f lamda = (60) (3) = 180 m//s" "...(8)`
(c) According to formula of velocity of transverse wave,
`v = sqrt ((T)/(mu)) = sqrt ((T)/(M //L)) = sqrt ((TL)/(M))`
`implies v ^(2) = (TL)/(M) implies T = (Mv ^(2))/( L )`
`therefore T = ((0.03) (180) ^(2))/((1.5)) =648 N" "...(9)`