# A long 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 2.56 m and mass 0.2 Kg. The wire PQR is under a tension of 80 N. 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 of the wire, and(b) the amplitude of the reflected and transmitted wave – pulses after the incident wave – pulse crosses the joint Q.

10 years ago
Hello Student,
(a) (Mass per unit length of PQ
m1 = 0.06/4.8 kg/m
Mass per unit length of QR, m2­ = 0.2/2.56 kg/m
Velocity of wave in PQ is
v1 = √T/m1 = √80/0.06/4.8 = 80 ms-1 [ ∵ T = 80 N given]
Velocity of wave in QR is
v2 = √T/m2 = √80/0.2 / 2.56 = 32 m/s
∴ Time taken for the wave to reach from P to R
= tPQ + tQR
= 4.8/80 + 2.56/32 = 0.14 s
(b) When the wave which initiates from P reaches Q (a denser medium ) then it is partly reflected and partly transmitted.
In this case the amplitude of reflected wave
Ar = (v2 – v1/v2 + v1) A1 . . . . . . . . . . . . . . . . . . . . . . (i)
Where Ai = amplitude of incident wave also amplitude of transmitted wave is
At = (2 v2/v1 + v2) Ai . . . . . . . . . . . . . . . . . . . . . . .(ii)
From (i), (ii)
Therefore, At = 2 cm and Ar = -1.5 cm

Thanks