A long wire PQR is made by joining two wires PQ and QR of equal radii. PQ has a length 4.8m and mass 0.06 kg. QR has length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80N. A sinusoidal wave pulse of amplitude 3.5 cm is sent along the wire PQ from the end P. No power is dissipated during the propagation of the wave pulse. Calculate:
(a) The time taken by the wave pulse to reach the other end R.
(b) The amplitude of the reflected and transmitted wave pulse after the incident wave pulse crosses the joint Q.

Answer

(i) ` 140 ms
(ii) `A_(r)=((v_(2)-v_(1))/(v_(2)+v_(1)))A_(1)=1.5cm,A_(t)=((2v_(2))/(v_(2)+v_(1)))A_(2)=2cm`

Explanation

Amplitude of incident wave
`A_(1) = 3.5 cm`

Tension `T = 80 N`
Mass per unit length of wire PQ is
`m_(1) = (0.06)/(4.8) = (1)/(80) kg//m`
and mass per unit length of wire QR is
`m_(2) = (0.2)/(2.56) = (1)/(12.8) kg//m`
(i) Speed of wave in wire PQ is
`V_(1)=sqrt((T)/(m_(1)))=sqrt((80)/(1//80))=80m//s`
and speed and wave in wire QR is
`V_(2)=sqrt((T)/(m_(2)))=sqrt((80)/(1//128))=32m//s`
`:.` Time taken by the wave pulse to reach from P to R is
`t=(4.8)/(V_(1))+(2.56)/(V_(2))=((4.8)/(80)+(2.56)/(32))srArr t=0.14s`
(ii) The expressions for reflected and transmitted amplitudes `(A_(r) and A_(l))` in terms of `V_(1),V_(2)` and `A_(1)` are as follpws :
`A_(r)=(V_(2)-V_(1))/(V_(2)+V_(1))A_(i)and A_(t) =(2V_(2))/(V_(1)+V_(2))A_(l)`
Substituting the values, we get
`A_(r)=((32-80)/(32+80))(3.5)=-1.5cm`
i.e., the amplitude of reflected wave will be 1.5 cm. Negative sign of `A_(r)` indicates that there will be a phase change of `pi` in reflected wave. Similarly.
`A_(t)=((2xx32)/(32+80))(3.5)=2.0`
i.e., the amplitude of transmitted wave will be 2.0 cm.