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 a length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80 N . A sinusoidal wave pulse of amplitude 3.5cm is sent along the wire PQ from the end? No power is dissipated during the propagation of the wave - pulse . Calculate (a) The time taken by the wave pulse to reach the order end R of the wire , and (b) The amplitude of the reflected and transmitted wave pulses after the incident wave pulse crosses the joint Q.

A long wire PQR is made by joining two wires PQ and QR of equal radii

1.	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 a length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80 N . A sinusoidal wave pulse of amplitude 3.5cm is sent along the wire PQ from the end? No power is dissipated during the propagation of the wave - pulse . Calculate (a) The time taken by the wave pulse to reach the order end R of the wire , and (b) The amplitude of the reflected and transmitted wave pulses after the incident wave pulse crosses the joint Q.
Answer» Solution :Given that `m_(1) = 0.06 kg, m_(2) = 0.2 kg` Let `m' ` be the mass PER unit LENGTH then `m'_(1) = 0.0125 kg//m , m'_(2) = 0.078125 kg//m` Wire `PQR` is under a TENSION of `80 N= T_(0).A` sinusoidal wave pulse is sent from `P`. (a) `v_(1)` = velocity of wave on `PQ` `= sqrt((T)/(m_(1))) = sqrt((80)/(0.0125)) = 80 m//s` `v_(2)` = velocity of wave on `QR` `sqrt((T)/(m_(2))) = sqrt((80)/( 0.078125)) =32m//s` Total time tken for wave pulse to reach from `P` to `R` ` = (PQ)/( v_(1)) + (QR) /(v_(2)) = (( 4.8)/( 80) + ( 2.56)/( 32)) s = 0.06 + 0.08 = 0.14 s` (b) Amplitude of reflected wave : `A_(r) = (( v_(2) - v_(1))/( v_(2) + v_(1))) A_(i)` ` = (( 32 - 80)/( 32 + 80)) 3.5 = - 1.5 CM` `A_(r) is -ve` , so reflected wave is inverted Amplitude of transmitted wave : `A_(t) = (( 2 v_(2))/( v_(2) + v_(1))) A_(i) = (( 2 xx 32)/( 32 + 80 )) 3.5 = 2 cm`