Question

A long wire $PQR$ is made by joining two wires $PQ$ and $QR$ of equal radii. $PQ$ has length $4.8m$ and mass $0.06kg$ and $QR$ has length $2.56m$ and mass $0.2kg$. The wire $PQR$ is under a tension of $80N$. A sinusoidal wave-pulse of amplitude $3.5cm$ 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 $Q$.

Answer 1

We know that,

Velocity = $\sqrt{\frac{T}{Y}}$

$\therefore Y_{PQ}=\frac{0.06}{4.8}=\frac{6}{\left(100\right)}\begin{array}{c}\left(10\right)\\ \left(48\right)\end{array}=\frac{1}{80}kg/m$

$\therefore Y_{QR}=\frac{0.2}{2.56}=\frac{2}{\left(10\right)}\begin{array}{c}\left(100\right)\\ \left(256\right)\end{array}=\frac{10}{128}kg/m$

$\therefore V\left(PQ\right)=\sqrt{\frac{80}{\frac{1}{80}}}-80m/s$

$\therefore V\left(QR\right)=\sqrt{\frac{80}{\frac{10}{128}}}=\sqrt{8\times 128}$

$=4\times 16=64m/s$

$\therefore$ Time = $\frac{4.8}{80}+\frac{2.56}{64}=0.06+0.04$

$=0.1sec$

b) we know that

$a_{reflected}=\frac{(v_2-v_1)}{(v_2+v_1)}{a}_i$

$=\left(\frac{16}{144}\right)a_i$

$\therefore a_{reflected}=\frac{a_i}{9}=\frac{3.5}{9}=0.4cm$

$\therefore a_{transmitted}=\left(\frac{2v_2}{v_1+v_2}\right)a_i$

$=\frac{2\times 64}{144}\left(a_i\right)=\frac{8}{9}\times 3.5=3.2cm$