Two wires of different linear mass densities are soldered together end to end and then stretched under a tension `F`. The wave speed in the first wire is thrice that in the second. If a harmonic wave travelling in the first wire is incident on the junction of the wires and if the amplitude of the incident wave is `A = sqrt(13) cm`, find the amplitude of reflected wave.

Answer

`(sqrt (13))/(2) cm`

Explanation

By conservation of energy , the incident power equals the transmitted power plus the reflected power
`P_( in) = P_(t) + P_(r)`
`(1)/(2) mu_(1) omega_(2) A_(1 n)^(2) v_(1) = (1)/(2) mu _(2) omega^(2) A_(t)^(2) v_(2) + (1)/(2) mu_(1) omega^(2) A_(r)^(2) v_(1)` `(i)`
` mu_(1) = F//v_(1)^(2) and mu_(2) = F//v_(2)^(2)` `(ii)`
From Eqs. (i) and (ii)
`(A_( 1n)^(2))/( v_(1)) = (A_(t)^(2))/( v_(2)) + (A_(r)^(2))/( v_(1))`
Given , `v_(1) = 3 v_(2)`
`A_(1 n)^(2) = 3 A_(t)^(2) + A_(r)^(2)` `(iii)`
`A_(t) = (( 2v_(2))/(v_(1) + v_(2))) A_( 1n) = (A_(1 n))/(2)`
`A_(r) = ((v_(2) - v_(1))/( v_(2) + v_(1))) A _( 1n) = ( -A_( 1n))/( 2)`
` = (- sqrt(13))/(2) cm`