Question

A heavy piece of hanging sculpture is suspended by a90-cm-long, 5.0 g steel wire. When the wind

blows hard, the wire hums at its fundamental frequency of 80Hz. What is the mass of the

sculpture?

Answer 1

Concepts and reason

The concepts required to solve this problem are linear density, tension in wire and fundamental frequency of the string.

To calculate the mass of the sculpture firstly calculate the tension in wire and then calculate the linear density.

Fundamentals

The general expression for fundamental frequency of the string is,

$f = \frac{1}{{2L}}\sqrt {\frac{T}{\mu }}$

Here, $L$is length of the wire , $\mu$is the linear density and $T$is the tension in the wire.

The expression for tension in wire is,

$T = Mg$

Here, $M$is mass of the sculpture and $g$is the acceleration due to gravity.

The expression for linear density is,

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

Here, $m$is mass of the wire.

The expression for linear density is,

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

Substitute $5{\rm{ g}}$ for $m$and $90\,\,{\rm{cm}}$ for$L$ in above equation of linear density.

$\begin{array}{c}\\\mu = \frac{{5{\rm{ g}}\left( {\frac{{1\;{\rm{kg}}}}{{{{10}^3}{\rm{ g}}}}} \right)}}{{{\rm{90 cm}}\left( {\frac{{1\;{\rm{m}}}}{{{{10}^2}\;{\rm{cm}}}}} \right)}}\\\\ = 5.555 \times {10^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^{ - 1}}\\\end{array}$

The expression for tension in wire is,

$T = Mg$

Substitute $9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for $g$in above equation tension.

$T = M\left( {9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)$

The general expression for fundamental frequency of the string is,

$f = \frac{1}{{2L}}\sqrt {\frac{T}{\mu }}$

Substitute $5.555 \times {10^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^{ - 1}}$ for $\mu$, $90\,\,{\rm{cm}}$ for$L$ and $M\left( {9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)$ for $T$in above equation.

$\begin{array}{c}\\f = \frac{1}{{2\left( {90\,\,{\rm{cm}}\left( {\frac{{1\;{\rm{m}}}}{{{{10}^2}\;{\rm{cm}}}}} \right)} \right)}}\sqrt {\frac{{M\left( {9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)}}{{5.555 \times {{10}^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^{ - 1}}}}} \\\\ = 0.555\;{{\rm{m}}^{ - 1}}\sqrt {\frac{{M\left( {9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)}}{{5.555 \times {{10}^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^{ - 1}}}}} \\\end{array}$

Substitute $80{\rm{ Hz}}$ for $f$ in above equation.

$\begin{array}{c}\\\frac{{80{\rm{ Hz}}}}{{0.555\;{{\rm{m}}^{ - 1}}}} = \sqrt {\frac{{M\left( {9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)}}{{5.555 \times {{10}^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^{ - 1}}}}\left( {1764{\rm{ k}}{{\rm{g}}^{ - 1}} \cdot {{\rm{m}}^2} \cdot {{\rm{s}}^{ - 2}}} \right)} \\\\{\left( {144{\rm{ Hz}} \cdot {\rm{m}}} \right)^2} = \frac{{M\left( {9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)}}{{5.555 \times {{10}^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^{ - 1}}}}\\\\M = 11.76{\rm{ kg}}\\\end{array}$

Ans:

The mass of the sculpture is $11.76{\rm{ kg}}$.