Physics - Wave on String Question with Solution | TestHub

The average power transmitted by a wave on a string is $P_{avg}=\frac{1}{2}\mu A^2{\omega }^{2}v$. Here, $\mu$ and $v=\sqrt{T/\mu }$ (since $T$ and $\mu$ are constant) remain constant. So, $P_{avg}\propto A^2{\omega }^2$.

Let the initial amplitude be $A_1=A$ and initial angular frequency be ${\omega }_1=\omega$. The initial power is $P_1=k{A}^2{\omega }^2$.

The amplitude is doubled, $A_2=2A$. The frequency is halved, so the angular frequency is also halved, ${\omega }_2=\frac{\omega }{2}$. The new power $P_2$ is:$$P_2=k(2A{)}^2(\frac{\omega }{2}{)}^2=k(4A^2)(\frac{{\omega }^2}{4})=k{A}^2{\omega }^2$$ Thus, $P_2=P_1=P$.