Question

​6.A long wire PQR is made by joining two wires PQ and QR of equal radii PQ has 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 wire PQ FROM THE end P. no power is dissipated during the propagation of the wave pulse . calculate the amplitude of the reflected and transmitted wave pulses after the incident wave pulse crosses the joint Q.

Answer 1

Dear student
$$\begin{gathered} Amplitudeofreflectedwaveis \\ {A}_r=\left(\frac{\sqrt{{\mu }_1}-\sqrt{{\mu }_2}}{\sqrt{{\mu }_1}+\sqrt{{\mu }_2}}\right)\times A_i({\mu }_{1}isthemassperunitlengthofthewirePQand{\mu }_{2}isthemassperunitlengthofthewireQR) \\ diidingbothnumeratoranddenomenatorby\sqrt{{\mu }_2} \\ {A}_r=\left(\frac{\sqrt{{\displaystyle \frac{{\mu }_1}{{\mu }_2}}}-1}{\sqrt{{\displaystyle \frac{{\mu }_1}{{\mu }_2}}}+1}\right)\times A_i \\ {\mu }_1=\frac{.06}{4.8}and{\mu }_2=\frac{0.2}{2.56} \\ \frac{{\mu }_1}{{\mu }_2}=\frac{\frac{.06}{4.8}}{\frac{0.2}{2.56}}=\frac{2}{5} \\ {A}_r=\left(\frac{{\displaystyle \frac{2}{5}}-1}{({\displaystyle \frac{2}{5}}+1)}\right)\times 3.5=\left(\frac{2-5}{2+5}\right)\times 3.5=\frac{-3}{7}\times 3.5=\frac{3}{2} \\ and \\ {A}_t=A_i+A_r \\ {A}_t=3+\frac{3}{2}=\frac{9}{2}=4.5cm \\ Regards \end{gathered}$$