Physics - Wave on String Question with Solution | TestHub

The average power transmitted by a sinusoidal transverse wave on a string is given by $P=\frac{1}{2}\mu A^2{\omega }^{2}v$, where $v=\sqrt{T/\mu }$ is the wave speed. Substituting$v$ into the power equation:$P=\frac{1}{2}\mu A^2{\omega }^2\sqrt{\frac{T}{\mu }}=\frac{1}{2}{A}^2{\omega }^2\sqrt{T\mu }$

The angular frequency$\omega =2\pi f$. Since $A$, $f$ (and thus $\omega$), and $T$ are kept constant,

the power depends only on the linear mass density $\mu$.

For the first string,$P_1=\frac{1}{2}{A}^2{\omega }^2\sqrt{T{\mu }_1}$.

The second string is made of the same material (same density$\rho$) but has twice the radius ($r_2=2r_1$). The linear mass density $\mu =\rho \times (\text{cross-sectional area})=\rho \pi r^2$.

So,${\mu }_2=\rho \pi r_2^2=\rho \pi (2r_1{)}^2=4\rho \pi r_1^2=4{\mu }_1$.

For the second string,$P_2=\frac{1}{2}{A}^2{\omega }^2\sqrt{T{\mu }_2}$.

Substituting${\mu }_2=4{\mu }_1$:$P_2=\frac{1}{2}{A}^2{\omega }^2\sqrt{T(4{\mu }_1)}=\frac{1}{2}{A}^2{\omega }^2(2\sqrt{T{\mu }_1})=2\left(\frac{1}{2}{A}^2{\omega }^2\sqrt{T{\mu }_1}\right)=2P_1$ Thus, the new average power transmitted is$2P_1$.