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`.

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`