#### 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.Then what is the $A_r=\text{amplitude of the reflected wave pulse }$ and $A_t=\text{amplitude of the transmitted wave pulse }$ after the incident wave pulse crosses the joint Q.  Option: 1 $A_r=1.5 \ cm \ \ and \ \ A_t= 2 \ cm$Option: 2 $A_r=2.5 \ cm \ \ and \ \ A_t= 2 \ cm$Option: 3 $A_r=1.5 \ cm \ \ and \ \ A_t= 2.5 \ cm$Option: 4 $A_r=-1.5 \ cm \ \ and \ \ A_t= 2.5 \ cm$

$\text {Amplitude of wave pulse }=3.5 \mathrm{~cm}$
$\\ \mu_{\mathrm{PQ}}=\frac{\mathrm{m}}{\mathrm{L}}=\frac{0.06}{4.8}=0.0125 \mathrm{~kg} / \mathrm{m}\\\\ \mu_{\mathrm{QR}}=\frac{\mathrm{m}}{\mathrm{L}}=\frac{0.2}{2.56}=0.078125 \mathrm{~kg} / \mathrm{m}$

The velocity of the wave in PQ wire
$V_1=\mathbf{V}_{PQ}=\sqrt{\frac{\mathbf{T}}{\mu_{\mathrm{P} Q}}}=\sqrt{\frac{80}{0.0125}}=\mathbf{8 0 m} / \mathbf{s}$

The velocity of the wave in QR length
$V_2=\mathbf{V}_{QR}=\sqrt{\frac{\mathbf{T}}{\mathbf{\mu}_{QR}}} =\sqrt{\frac{\mathbf{8 0 }}{0.078125}}=32 \ m/s$

And The expressions for reflected and transmitted amplitudes $(A_{r} \ and \ A_{t})$ in terms of $V_{1}, V_{2} \ and \ A_{i}$ are as follows:
$A_{r}= \frac{\left ( V_{2}-V_{1} \right )A_{i}}{\left ( V_{2}+V_{1} \right )}$ and  $A_{t}= \left ( \frac{2V_{2}}{V_{1}+V_{2}} \right )A_{i}$

So using this we get

$\mathrm{A_{r}=\left(\frac{32-80}{32+80}\right)(3.5)=-1.5 c m}$

i.e. the amplitude of the reflected wave will be 1.5 cm.

The negative sign of  $A_r$ indicates that there
will be a phase change of $\pi$ in the reflected wave.

$\text { Similarly, } A_{t}=\left(\frac{2 \times 32}{32+80}\right)(3.5)=2.0 \mathrm{~cm}$

i.e. the amplitude of transmitted wave will be
2.0 cm.