Question

A $400\text{Hz}$ wave with amplitude $2\text{mm}$ travels on a long string of linear mass density $5\text{g/m}$ kept under a tension of $50\text{N}$. Find the average power and total energy associated with the wave in a $2\text{m}$ long portion of the string.

• A. $6\text{W}$ , $1\text{J}$
• B. $1\text{W}$ , $6\text{J}$
• C. $6.32\text{W}$ , $0.13\text{J}$
• D. None of the above

Answer 1

The correct option is C $6.32\text{W}$ , $0.13\text{J}$

Given that
$f=400\text{Hz},A=2\text{mm}=2\times {10}^{-3}\text{m}$
$\mu =5\text{g/m}=5\times {10}^{-3}\text{kg/m}$
$T=50\text{N}$ and length $\left(l\right)=2\text{m}$

Average Power $\left(P\right)=2{\pi }^2{f}^2{A}^2\mu v$
$=2{\pi }^2{f}^2{A}^2\sqrt{T\mu }[\because v=\sqrt{\frac{T}{\mu }}]$
$=2{\pi }^2\times (400{)}^2\times (2\times {10}^{-3}{)}^2\times \sqrt{50\times 5\times {10}^{-3}}$
$=6.32\text{W}$

And, Energy density $U$ = $\frac{1}{2}\rho {\omega }^2{A}^2$
Total energy = $U\times volume$
$=\frac{1}{2}\times (2\pi f{)}^2\times A^2\times \mu \times l$
[$\because \mu \times l=$ total mass = volume $\times \rho$]
$=\frac{1}{2}\times (2\pi \times 400{)}^2\times (2\times {10}^{-3}{)}^2\times 5\times {10}^{-3}\times 2$
$=0.1263\text{J}$
$\approx 0.13\text{J}$