Question

A wave moving with constant speed on a uniform string passes the point x=0 with amplitude $A_0$,angular frequency ${\omega }_o$ and average rate of energy transfer $P_0$.As the wave travels down the string it gradually loses energy and at the point x = l, the average rate of energy transfer becomes $\frac{P_0}{2}$. At the point x = l, angular frequency and amplitude are respectively:

• A. ${\omega }_0$ and $A_0/\sqrt{2}$
• B. ${\omega }_0/\sqrt{2}$ and $A_0$
• C. less than ${\omega }_0$ and $A_0$
• D. ${\omega }_0/\sqrt{2}$ and $A_0/\sqrt{2}$

Answer 1

The correct option is A ${\omega }_0$ and $A_0/\sqrt{2}$

$\begin{array}{cc}\text{Given}-& \ast \text{Initial Amplitude}=A_0\\ & \ast \text{Initial Angular trequency}={\omega }_0\\ & \ast \text{Rate of Power Transfer}=P_0\end{array}$

$\begin{array}{c}\text{We know that average Rate of Power Transfer (P)}\\ \begin{array}{cc}P\alpha A^2& A\to \text{Amplitude of wave}\\ \Rightarrow \frac{P_i}{P_f}=\frac{A_i^2}{A_f^2}& A_{+}\to \text{Final Amplitude}\\ \Rightarrow \frac{P_0}{P_0/2}=\frac{A_0^2}{A_f^2}& \\ \Rightarrow A_f=\frac{A_0}{\sqrt{2}}& \end{array}\end{array}$

$\begin{array}{c}\text{And angular frequency of the wave always Remain}\\ \text{constant}\\ \Rightarrow w_{+}={\omega }_0\text{and}{A}_t=A_0/\sqrt{2}\\ \text{Hence, option (A) is correct.}\end{array}$