Question

A sinusoidal wave of amplitude $5\text{cm}$ is to be transmitted along a string having a linear mass density of $4\times {10}^{-2}\text{kg/m}$. If the source can be deliver an average power of $90\text{watts}$ and string is under a tension of $100\text{N}$, then the frequency at which the source operates is
[Take ${\pi }^2=10$]

• A. $45.3\text{Hz}$
• B. $50\text{Hz}$
• C. $30\text{Hz}$
• D. $62.3\text{Hz}$

Answer 1

The correct option is C $30\text{Hz}$

Average power $\left(P\right)=\frac{1}{2}\mu {\omega }^2{A}^{2}v$
As per question :
$T=100\text{N}$
$A=5\text{cm}=5\times {10}^{-2}\text{m}$
$\mu =4\times {10}^{-2}\text{kg/m}$
$P=90\text{watts}$

Wave velocity $\left(v\right)=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{100}{4\times {10}^{-2}}}=\frac{100}{2}=50\text{m/s}$
So, $P=\frac{1}{2}\mu {\omega }^2{A}^{2}v$
$\Rightarrow \omega =\sqrt{\frac{2P}{\mu A^{2}v}}=\sqrt{\frac{2\times 90}{4\times {10}^{-2}\times (5\times {10}^{-2}{)}^2\times 50}}$
$\Rightarrow \omega =\sqrt{36000}\text{rad/s}$

$\omega =2\pi f$
$\Rightarrow f=\frac{\omega }{2\pi }=\sqrt{\frac{36000}{2\pi }}=\frac{60\sqrt{10}}{2\sqrt{10}}=30\text{Hz}$
[using ${\pi }^2=10$]