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.

Explanation

The tension is maintained at 80N and the amplitude of the incident wave is given to be 3.5 cm
The mass per unit length of `PQ=mu_(1)=(0.06)/(4.8)=(1)/(80)" kg "m^(-1)`

And the mass per unit length of `QR=(0.2)/(2.56)=(1)/(12.8)" kg "m^-1`
speed `v_(1)` in `PQ=sqrt((T)/(mu_(1)))=sqrt((80)/(1//80))=80ms^(-1)` and speed `v_(2)` in `QR=sqrt((T)/(mu_(2)))=sqrt((80)/(1//12.8))=32ms^(-1)`
(i) Time taken by th wave pulse to move from P to R is given by
`t=(l_(1))/(v_(1))+(l_(2))/(v_(2))=(4.8)/(80)+(2.56)/(32)=0.14s`
(ii) `A_(r)=((V_(2)-V_(1))/(V_(2)+V_(1)))A_(i)=((32-80)/(32+80))xx3.5=-1.5cm`
`A_(t)=((2V_(2))/(V_(2)+V_(1)))A_(i)=(2xx32)/(32+80)3.5=2cm`