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Question

# 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 and 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-pulse after the incident wave pulse cross the joint Q.

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Solution

## We know that, Velocity = √TY∴YPQ=0.064.8=6(100)(10)(48)=180kg/m∴YQR=0.22.56=2(10)(100)(256)=10128kg/m∴V(PQ)=  ⎷80180−80m/s∴V(QR)=  ⎷8010128=√8×128 =4×16=64m/s∴ Time = 4.880+2.5664=0.06+0.04 =0.1secb) we know that areflected=(v2−v1)(v2+v1)ai =(16144)ai∴areflected=ai9=3.59=0.4cm∴atransmitted=(2v2v1+v2)ai=2×64144(ai)=89×3.5=3.2cm

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