A long wire `PQR` is made by joining two wires `PQ` and `QR` of equal radii. `PQ` has length `4.8 m` and mass `0.06 kg`. `QR` has length `2.56m` and mass `0.2 kg`. The wire `PQR` is under a tension of `80N`. A sinusoidal wave-pulse of amplitude `3.5cm` is sent along the wire `PQ` from 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` of the wire, and
(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.