Question

A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplitude 1.6 mm travels along the wire. Calculate the average power carried by the wave. What is the average power if the wave amplitude is halved?

Answer 1

Concept and reason

The concept needed to solve this problem is average power dissipated by a wave on a string.

Initially, write the expression for the average power carried by a wave on a string. Later, write the relation between the angular frequency and frequency. Use this relation in the previous equation. After that, write equation to calculate speed of a wave on a string and use this equation first equation. Write the expression for the linear mass density of the string by doing mass divided by the length of the string. Finally, derive an expression for the average power carried by a wave on a string in terms of all known quantities.

Fundamentals

The expression for the average power carried by a wave on a string is,

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

Here, $\mu$ is the linear mass density of the string, $\omega$ is the angular frequency of the wave on the string, $A$ is the amplitude of the wave, and v is the speed of the wave.

The expression to calculate the speed of a wave on a string is,

$v = \sqrt {\frac{T}{\mu }}$

The linear mass density of the string can be calculated using the following equation:

$\mu = \frac{m}{l}$

Here, m is the mass of the string and l is the length of the string.

The relation between the angular frequency and linear frequency (f) is,

$\omega = 2\pi f$

Substitute $\sqrt {\frac{T}{\mu }}$ for v, $\frac{m}{l}$ for $\mu$ , $2\pi f$ for $\omega$ in the equation $P = \frac{1}{2}\mu {\omega ^2}{A^2}v$ .

$\begin{array}{c}\\P = \frac{1}{2}\left( {\frac{m}{l}} \right){\left( {2\pi f} \right)^2}{A^2}\left( {\sqrt {\frac{T}{\mu }} } \right)\\\\ = \frac{1}{2}\left( {\frac{m}{l}} \right){\left( {2\pi f} \right)^2}{A^2}\left( {\sqrt {\frac{T}{{\left( {\frac{m}{l}} \right)}}} } \right)\\\\ = 2{\pi ^2}{f^2}{A^2}\left( {\sqrt {T\left( {\frac{m}{l}} \right)} } \right)\\\end{array}$

(A)

The final equation for average power carried by the wave is,

$P = 2{\pi ^2}{f^2}{A^2}\left( {\sqrt {T\left( {\frac{m}{l}} \right)} } \right)$

Convert the unit of mass of the string from gram into kilogram:

$\begin{array}{c}\\m = 2.60{\rm{ g}}\left( {\frac{{1{\rm{ kg}}}}{{{{10}^3}{\rm{ g}}}}} \right)\\\\ = 2.60 \times {10^{ - 3}}{\rm{ kg}}\\\end{array}$

Convert the unit of length of string from centimeter to meter:

$\begin{array}{c}\\l = 84.0{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\\\\ = 0.84{\rm{ m}}\\\end{array}$

Convert the unit of amplitude from millimeter to meter:

$\begin{array}{c}\\A = 1.6{\rm{ mm}}\left( {\frac{{1{\rm{ m}}}}{{{{10}^3}{\rm{ mm}}}}} \right)\\\\ = 1.6 \times {10^{ - 3}}{\rm{ m}}\\\end{array}$

Substitute $2.60 \times {10^{ - 3}}{\rm{ kg}}$ for m, $0.84{\rm{ m}}$ for l, $1.6 \times {10^{ - 3}}{\rm{ m}}$ for A, 25.0 N for T, and 120.0 Hz in the equation $P = 2{\pi ^2}{f^2}{A^2}\left( {\sqrt {T\left( {\frac{m}{l}} \right)} } \right)$ .

$\begin{array}{c}\\P = 2{\pi ^2}{\left( {120.0{\rm{ Hz}}} \right)^2}{\left( {1.6 \times {{10}^{ - 3}}{\rm{ m}}} \right)^2}\left( {\sqrt {\left( {25.0{\rm{ N}}} \right)\left( {\frac{{2.60 \times {{10}^{ - 3}}{\rm{ kg}}}}{{0.84{\rm{ m}}}}} \right)} } \right)\\\\ = {\rm{0}}{\rm{.2024 W}}\\\end{array}$

[Part A]

(B)

The new amplitude $A'$ that is half of the wavelength of the wave is,

$\begin{array}{c}\\A' = \frac{{1.6 \times {{10}^{ - 3}}{\rm{ m}}}}{2}\\\\ = 0.8 \times {10^{ - 3}}{\rm{ m}}\\\end{array}$

The final equation for average power carried by the wave is,

$P = 2{\pi ^2}{f^2}{A^2}\left( {\sqrt {T\left( {\frac{m}{l}} \right)} } \right)$

Substitute $2.60 \times {10^{ - 3}}{\rm{ kg}}$ for m, $0.84{\rm{ m}}$ for l, $0.8 \times {10^{ - 3}}{\rm{ m}}$ for A, 25.0 N for T, and 120.0 Hz. $\begin{array}{c}\\P = 2{\pi ^2}{\left( {120.0{\rm{ Hz}}} \right)^2}{\left( {0.8 \times {{10}^{ - 3}}{\rm{ m}}} \right)^2}\left( {\sqrt {\left( {25.0{\rm{ N}}} \right)\left( {\frac{{2.60 \times {{10}^{ - 3}}{\rm{ kg}}}}{{0.84{\rm{ m}}}}} \right)} } \right)\\\\ = {\rm{0}}{\rm{.0506 W}}\\\end{array}$

[Part B]

Ans: Part A

The average power carried by the wave is 0.2024 W.

Part B

The new average power carried by the wave is 0.0506 W.