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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

Answers (1)


\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.

Posted by

Ajit Kumar Dubey

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