Question

A $200Hz$ wave with amplitude $1mm$ travels on a long string of linear mass density $6g{m}^{-1}$ kept under a tension of $60N$.

(a) Find the average power transmitted across a given point on the string.

(b) Find the total energy associated with the wave in a $20m$ long portion of the string.

Answer 1

Step 1: Given data

Amplitude of the wave, $A=1mm$

$={10}^{-3}m\left[As1m={10}^{3}mm\right]$

Frequency of the wave, F$=200Hz$

Linear mass density, m$=6g{m}^{-1}$

Tension in the string, $T=60N$

Step 2: Find the velocity of the wave

Velocity of the wave, $v=\sqrt{T/m}$

$$\begin{gathered} =\sqrt{\left(60/6\times {10}^{-3}\right)} \\ =100m/s \end{gathered}$$

Step 3:(a) Find the transmitted power

Average transmitted power, $P=2{\pi }^2\times m\times v\times A^2\times f^2$

$$\begin{gathered} =2\times {\left(3.14\right)}^2\times 6\times {10}^{-3}\times 100\times {\left({10}^{-3}\right)}^2\times {\left(200\right)}^2 \\ =0.47W \end{gathered}$$

Step 4: Find the time required to cover $\mathbf{20}\mathbf{}\mathit{m}$ distance

Time required to cover $20m$distance $=\left(20/100\right)s$

$=0.2s$

Step 5:(b) Find the total energy associated with $\mathbf{20}\mathbf{}\mathit{m}$

Energy associated with $20m$ string, $E=2{\pi }^2\times m\times v\times t\times A^2\times f^2$

$$\begin{gathered} =2\times {\left(3.14\right)}^2\times 6\times {10}^{-3}\times 100\times 0.2\times {\left({10}^{-3}\right)}^2\times {\left(200\right)}^2 \\ =0.094J\left[As1J={10}^{3}mJ\right] \\ =94mJ \end{gathered}$$

Hence,

(a) The average power transmitted across a given point on the string, $P=0.47W$

(b) Energy associated with $20m$ string, $E=94mJ$