# Power is to be transmitted along a taut string by means of transverse harmonic waves. The wave...

## Question:

Power is to be transmitted along a taut string by means of transverse harmonic waves. The wave speed is {eq}9\ \rm{m/s} {/eq} and the linear mass density of the string is {eq}0.009\ \rm{kg/m} {/eq}. The power source oscillates with an amplitude of {eq}0.46\ \rm{mm} {/eq}.

(a) What average power is transmitted along the string if the frequency is {eq}420\ \rm{Hz} {/eq}? Answer in {eq}\rm{mW} {/eq}.

(b) The power transmitted can be increased by increasing the tension in the string, the frequency of the source, or the amplitude of the waves. By how much would each of these quantities have to increase to cause an increase in power by a factor of {eq}100 {/eq} if it is the only quantity changed?

## Power Transmission:

The power which is transmitted from one point to another depends upon the linear mass density, frequency, amplitude, and speed of wave in the string. Mathematically, the average power can be given as,

{eq}{P_a} = \dfrac{1}{2}\mu v{\omega ^2}{A^2} {/eq}

Here, {eq}{P_a} {/eq} is the average power, {eq}\mu {/eq} is the linear mass density, {eq}v {/eq} is the speed of wave, {eq}\omega {/eq} is the angular frequency and {eq}A {/eq} is the amplitude of wave.

## Answer and Explanation: 1

**Given Data:**

- The speed of wave is, {eq}v = 9\;{\rm{m/s}} {/eq}.

- The linear mass density of string is, {eq}\mu = 0.009\;{\rm{kg/m}} {/eq}.

- The amplitude of oscillation is, {eq}A = 0.46\;{\rm{mm}} {/eq}.

- The frequency of the transmission is, {eq}f = 420\;{\rm{Hz}} {/eq}.

**Part (a)**

The angular frequency of the transmission of signal is,

{eq}\omega = 2\pi f {/eq}

Plug in the known value,

{eq}\begin{align*} \omega &= 2\left( {3.14} \right)\left( {420\;{\rm{Hz}}} \right)\\ &= 2637.6\;{\rm{Hz}} \end{align*} {/eq}

The average power transmitted along the string can be expressed as,

{eq}{P_a} = \dfrac{1}{2}\mu v{\omega ^2}{A^2} {/eq}

Substituting known values,

{eq}\begin{align*} {P_a} &= \dfrac{1}{2}\left( {0.009\;{\rm{kg/m}}} \right)\left( {9\;{\rm{m/s}}} \right){\left( {2637.6\;{\rm{Hz}}} \right)^2}{\left( {\dfrac{{1\;{{\rm{s}}^{ - 1}}}}{{1\;{\rm{Hz}}}}} \right)^2}{\left( {0.46\;{\rm{mm}}} \right)^2}{\left( {\dfrac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)^2}\left( {\dfrac{{1\;{\rm{W}}}}{{1\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{3}}}}}} \right)\left( {\dfrac{{1\;{\rm{mW}}}}{{{{10}^{ - 3}}\;{\rm{W}}}}} \right)\\ &= 59.6\;{\rm{mW}} \end{align*} {/eq}

Thus, the average power of the signal transmission is {eq}56.9\;{\rm{mW}} {/eq}.

**Part (b)**

The speed of signal in the string is given as,

{eq}v = \sqrt {\dfrac{T}{\mu }} {/eq}

Therefore, the average power becomes,

{eq}\begin{align*} {P_a} &= \dfrac{1}{2}\mu \left( {\sqrt {\dfrac{T}{\mu }} } \right){\left( {2\pi f} \right)^2}{A^2}\\ &= 2\pi^{2} \sqrt {T\mu } {f^2}{A^2} \end{align*} {/eq}

Therefore, the power depends upon the square root of tension, square of frequency and square of amplitude.

If the tension of the string increased by factor {eq}25 {/eq} then the power will be increased by factor {eq}5 {/eq}, if the frequency is changed by factor {eq}2 {/eq} then the power will be increased by factor {eq}4 {/eq} and the amplitude can be increased by factor {eq}\sqrt 5 {/eq} to make the power increased by {eq}5 {/eq}, these factors will give the final power {eq}100P {/eq}.

Thus, the tension should be increased by {eq}25 {/eq}, frequency should be increased by {eq}2 {/eq} and amplitude should be increased by {eq}\sqrt 5 {/eq}.

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