Question

A long wire PQR is made by joining two wires PQ and QR of equal radii. PQ has a 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. The time taken by the wave pulse to reach the other end R is $14\times {10}^{-x}s$. Find x.

Answer 1

Amplitude of wave pulse=$3.5$cm

${\mu }_{PQ}=\frac{m}{L}=\frac{0.06}{4.8}=0.0125kg/m{\mu }_{QR}=\frac{m}{L}=\frac{0.2}{2.56}=0.0781kg/m$

Velocity of the wave in PQ wire

$v=\sqrt{\frac{T}{{\mu }_{PQ}}}=\sqrt{\frac{80}{0.0125}}=80m/s$

Time taken to travel PQ length

$t_{PQ}=\frac{d}{v}=\frac{4.8}{80}=0.06$sec

Velocity of the wave in QR length

$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{80+0.06\times 10}{0.0781}}=\sqrt{\frac{80.6}{0.0781}}$

v=$32.12$m/s

Time taken to travel QR length

$t_{QR}=\frac{d}{v}=\frac{2.56}{32.12}=0.079\approx 0.08$sec

Total time taken for the pulse to reach

t=$0.06+0.08$

t=$0.14$sec

t=$14\times {10}^{-2}sec\therefore x=2$

Hence the correct answer is $2$.