Power Transmitted Along The String

Power Transmitted Along the String

As a sinusoidal wave moves down a string, the energy associated with one wavelength on the string is transported down the string at the propagation velocity v. From the basic wave relationship the distance travelled in one period is vT = λ, so the energy is transported one wavelength per period of the oscillation.

The energy associated with one wavelength of the wave is

$E_{\lambda }=\frac{1}{2}\mu {\omega }^2{A}^2\lambda$

so the power transmitted would be :

$\begin{array}{rl}& P_{\lambda }=\frac{1}{2}\mu {\omega }^2{A}^2\frac{\lambda }{T}\\ & \text{since}v=\frac{\lambda }{T}\\ & \text{Therefore}{P}_{\lambda }=\frac{1}{2}\mu {\omega }^2{A}^{2}v\end{array}$

where $\omega =$ angular frequency, $\mu =$ mass per unit length of string,$A=$ wave amplitude $v=$ wave propogation velocity

The Intensity of the Wave

The intensity of a wave refers to the amount of energy the wave transmits per unit area perpendicular to the direction of propagation, typically measured in watts per square meter (W/m²). It is a crucial parameter that determines the strength and impact of a wave, whether it's sound, light, or any other type of wave. The intensity is directly proportional to the square of the wave's amplitude, meaning that even small increases in amplitude can lead to significant increases in intensity.

The flow of energy per unit area of the cross-section of the string in the unit time is known as the intensity of the wave.

${}_{\mathrm{As}}P=\frac{1}{2}\mu {\omega }^2{A}^{2}v$

And using $I=\frac{P}{\text{Area}}$
we get $I=\frac{\frac{1}{2}\mu {\omega }^2{A}^{2}v}{\text{Area}}$
using $\mu =\frac{\text{mass}}{\text{lenth}}=\frac{m}{l}$ and Volume $=$ Area $\times$ length
We get $I=\frac{\frac{1}{2}m{\omega }^2{A}^{2}v}{\text{lenth}\times \text{Area}}=\frac{\frac{1}{2}m{\omega }^2{A}^{2}v}{\text{Volume}}$
And now using $\rho =\frac{\text{mass}}{\text{volume}}$
we get $I=\frac{1}{2}\rho {\omega }^2{A}^{2}v$
Where
$\begin{array}{rl}& \rho =\text{density}\\ & \omega =\text{angular frequency}\\ & A=\text{Amplitude}\\ & \mathrm{v}=\text{Wave speed}\end{array}$

Solved Example Based On Power Transmitted Along The String

Example 1: A stretched rope has linear mass density $5\times {10}^{-2}\mathrm{kg}/\mathrm{m}$ and is under tension of 80 N. The power that has to be supplied (in W) to generate a wave of frequency 60 Hz and amplitude of $\frac{2\sqrt{2}}{15\pi }m$ is

1) 512

2) 251

3) 215

4) 521

Solution:

Rate of energy transfer on a string by a sinusoidal wave

$P=\frac{1}{2}\mu {\omega }^2{A}^{2}v$
wherein
$\mu =$ linear mass density
$\omega =$ angular frequency
$A=$ Amplitude
$v=$ Wave speed
$P=\frac{1}{2}\mu {\omega }^2{A}^2$ v where $v=\sqrt{\frac{T}{\mu }}$
After solving, $P=512\mathrm{W}$

Hence, the answer is the option (1).

Example 2: The prong of an electrically operated tuning fork is connected to a long string of $\mu =1\mathrm{kg}/\mathrm{m}$ and tension 25 N. The max velocity of the prong is 1 cm/s, then the average power needed to drive the prong is

1) $5\times {10}^{-4}\mathrm{W}$
2) $2.5\times {10}^{-4}\mathrm{W}$
3) ${10}^{-4}\mathrm{W}$
4) ${10}^{-3}\mathrm{W}$

Solution:

Rate of energy transfer on a string by a sinusoidal wave

$P=\frac{1}{2}\mu {\omega }^2{A}^{2}v$
wherein
$\begin{array}{rl}\mu & =\text{linear mass density}\\ \omega & =\text{angular frequency}\\ A& =\text{Amplitude}\\ v& =\text{Wave speed}\\ v& =\sqrt{\frac{F}{\mu }}=5\mathrm{m}/\mathrm{s}\\ P& =\frac{1}{2}\mu v(\omega A{)}^2=\frac{1}{2}\times 1\times 5{(1\times {10}^{-2})}^2=2.5\times {10}^{-4}\mathrm{W}\end{array}$

Hence, the answer is the option (2).

Example 3: Sinusoidal waves 5 cm in amplitude are to be transmitted along a string having a linear mass density equal to $4\times {10}^{-2}\mathrm{kg}/\mathrm{m}$. If the source delivers a maximum power of 90 W and tension is 100N, then the highest frequency ( in Hz) at which the source can operate is

1) 30

2) 50

3) 45

4) 62

Solution:

$\begin{array}{rl}& P=\frac{1}{2}\mu {\omega }^2{A}^{2}v,v=\sqrt{\frac{T}{\mu }}\\ & P=\frac{1}{2}{\omega }^2{A}^2\sqrt{T/\mu }\\ & f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{2P}{A^2\sqrt{T/\mu }}}=30\mathrm{Hz}\end{array}$

Hence, the answer is the option (1).

Example 4:

If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will be

1) Increased by a factor of 2

2) Decreased by a factor of 2

3) Decreased by a factor of 4

4) unchanged

Solution

The intensity of the wave

$\begin{array}{rl}& I=\frac{1}{2}\rho {\omega }^2{A}^{2}v\\ & \rho =\text{mass density}\\ & \omega =\text{angular frequency}\\ & A=\text{Amplitude}\\ & v=\text{Wave speed}\\ & I=2{\pi }^2{a}^2{n}^{2}v\rho \Rightarrow I\alpha a^2{n}^2\\ & \Rightarrow \frac{I_1}{I_2}={\left(\frac{a_1}{a_2}\right)}^{2}X{\left(\frac{n_1}{n_2}\right)}^2={\left(\frac{1}{2}\right)}^{2}X{\left(\frac{1}{1/4}\right)}^2\\ & \Rightarrow I_2=\frac{I_1}{4}\end{array}$

Hence, the answer is the option (3).

Example 5: A string 9 m long and fixed at its ends is driven by a 270 Hz vibrator. The string vibrates in its Fifth harmonic mode. The speed of the wave and its fundamental frequency

1) 54Hz

2) 56Hz

3) 57Hz

4) 58Hz

Solution

$\begin{array}{rl}& L=9m\\ & n=270Hz\end{array}$

$\begin{array}{rl}& f^{\prime }=5f_0\\ & 270=5f_0\\ & f_0=\frac{270}{5}\end{array}$

Hence, the answer is the option (1).

Summary
The power transmitted along a string and the intensity of a wave are key concepts in wave dynamics, influencing how energy is transferred through mediums like strings or cables. The relationship between wave parameters—such as tension, amplitude, and frequency—directly affects the energy and intensity of the wave. By understanding and applying these principles, one can optimize systems for efficient energy transmission, as demonstrated in various solved examples.