Question

A long wire $PQR$ is made by joining two wires $PQ$ and $QR$ of equal radii. $PQ$ has length $4.8m$ and mass $0.06kg$. $QR$ 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 the time taken (in $s$) by the wave pulse to reach the other end $R$ of the wire.

Answer 1

We know that: $v=\sqrt{\frac{T}{\mu }}$
Speed of transverse wave in a stretched string.
$v=\sqrt{\frac{T}{\mu }}$$=\sqrt{\frac{T}{\frac{M}{L}}}$
$T$: tension in the string,
$M$: mass of string,
$L$: length of string,

Wire $PQ$: $T=80N$, $M_1/L_1=\frac{0.06}{4.8}$
Speed of wave in $PQ$,
$v_1=\sqrt{\frac{T}{\frac{M_1}{L_1}}}$

$v_1=\sqrt{\frac{80}{\frac{0.06}{4.8}}}$

$=\sqrt{\frac{80\times 4.8}{0.06}}$

$=80m/s$

Wire $QR$: $T=80N$, $M_2/L_2=\frac{0.2}{2.56}$
Speed of wave in $QR$,
$v_2=\sqrt{\frac{T}{\frac{M_2}{L_2}}}$
$v_2=\sqrt{\frac{80}{\frac{0.2}{2.56}}}$

$=\sqrt{\frac{80\times 2.56}{0.2}}$

$=32m/s$

Time taken by wave pulse to move form P to R
$T=\frac{L_1}{v_1}+\frac{L_2}{v2}=$$\frac{4.8}{80}+\frac{2.56}{32}$$=0.06+0.08=0.14s$

Final answer: $\left(0.14\right)$