Chapter 23, Problem 38SP

A bar of dimensions $1.00{\text{ cm}}^{\text{2}}\times 200\text{ cm}$ and mass 2.00 kg is clamped at its center. When vibrating longitudinally, it emits its fundamental tone in unison with a tuning fork making 1000 vibration/s. How much will the bar be elongated if, when clamped at one end, a stretching force of 980 N is applied at the other end? [Hint: Look at Problem 22.11 and Chapter 12.]

To determine

The elongation in a bar when clamped at one end and a stretching force of $980\text{ N}$ is applied at the other end.

Answer to Problem 38SP

Solution:

$0.123\text{ mm}$

Explanation of Solution

Given data:

The dimensions of bar is $1.00{\text{ cm}}^2\times 200\text{ cm}$.

The mass of rod is $2.00\text{ kg}$.

The frequency of fundamental tone of longitudinal vibration is $1000\text{ vib/s}$.

The force applied at the end of the bar is $980\text{ N}$.

Formula used:

Write the expression of density of rectangular bar.

$\rho =\frac{m}{AL}$

Here, $\rho$ is the density of bar, $m$ is the mass of bar, $A$ is the area of cross section, and $L$ is the length of bar.

Write the expression of speed of sound in a rectangular bar.

$v=\sqrt{\frac{Y}{\rho }}$

Here, $v$ is the speed of sound in rectangular bar and $Y$ is the Young’s Modulus.

Write the expression of speed of longitudinal waves.

$v=\lambda f$

Here, $v$ is the speed of longitudinal waves, $\lambda$ is the wavelength of wave, and $f$ is the frequency of wave.

Write the expression of Young’s Modulus.

$Y=\frac{F{L}_o}{A\Delta L}$

Here, $Y$ is the Young’s Modulus, $F$ is the force, $L_o$ is the original length of bar, $A$ is the area of cross section, and $\Delta L$ is the change in length of bar.

Explanation:

When the bar is clamped at its center and vibrates longitudinally, its ends are free. Therefore, the bar must have antinodes at its ends and a node at its center. The distance between node and antinode is $\frac{\lambda }{4}$.

So, the expression for length of bar in terms of wavelength is,

$\begin{array}{c}L=2\left(\frac{\lambda }{4}\right)\\ =\frac{\lambda }{2}\end{array}$

Solve for $\lambda$.

$\lambda =2L$

Substitute $200\text{ cm}$ for $L$

$\begin{array}{c}\lambda =2\left(200\text{cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)\\ =2\left(2\text{ m}\right)\\ =4\text{ m}\end{array}$

The expression of speed of longitudinal waves is,

$v=\lambda f$

Substitute $4\text{ m}$ for $\lambda$ and $1000\text{ vib/s}$ for $f$

$\begin{array}{c}v=\left(4\text{ m}\right)\left(1000\text{ vib/s}\right)\\ =4000\text{ m/s}\end{array}$

The expression of density of rectangular bar is,

$\rho =\frac{m}{Al}$

Substitute $2.00\text{kg}$ for $m$, $1.00{\text{ cm}}^2$ for $A$, and $200\text{ cm}$ for $l$

$\begin{array}{c}\rho =\frac{\left(2.00\text{}\text{kg}\right)}{\left(1.00{\text{ cm}}^2\left(\frac{{10}^{-4}{\text{ m}}^2}{1{\text{ cm}}^2}\right)\right)\left(200\text{cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)}\\ =\frac{2.00\text{kg}}{\left({10}^{-4}{\text{ m}}^2\right)\left(2\text{m}\right)}\\ ={10}^4{\text{ kg/m}}^3\end{array}$

The expression of speed of sound in a rectangular bar is,

$\begin{array}{c}v=\sqrt{\frac{Y}{\rho }}\\ v^2=\frac{Y}{\rho }\end{array}$

Solve for $Y$.

$Y=\rho v^2$

Substitute ${10}^4{\text{ kg/m}}^3$ for $\rho$ and $4000\text{ m/s}$ for $v$

$\begin{array}{c}Y=\left({10}^4{\text{kg/m}}^3\right){\left(4000\text{m/s}\right)}^2\\ =1.6\times {10}^{-11}\text{ Pa}\end{array}$

Also, the expression of Young’s Modulus is,

$Y=\frac{F{L}_o}{A\Delta L}$

Solve for $\Delta L$.

$\Delta L=\frac{F{L}_o}{AY}$

Substitute $980\text{ N}$ for $F$, $200\text{cm}$ for $L_o$, $1.00{\text{ cm}}^2$ for $A$, and $1.6\times {10}^{11}\text{ Pa}$ for $Y$

$\begin{array}{c}\Delta L=\frac{\left(980\text{N}\right)\left(200\text{ cm}\left(\frac{{10}^{-2}\text{m}}{1\text{ cm}}\right)\right)}{\left(1.00{\text{ cm}}^2\left(\frac{{10}^{-4}{\text{m}}^2}{1{\text{ cm}}^2}\right)\right)\left(1.6\times {10}^{11}\text{Pa}\right)}\\ =\frac{\left(980\right)\left(2\right)}{\left({10}^{-4}\right)\left(1.6\times {10}^{11}\right)}\text{ m}\\ \text{=}\frac{1960}{1.6\times {10}^7}\text{m}\\ \text{=1}\text{.23}\times {\text{10}}^{-4}\text{m}\end{array}$

Further, simplify the expression.

$\begin{array}{c}\Delta L=\left(0.123\times {10}^{-3}\text{ m}\right)\left(\frac{1\text{mm}}{{10}^{-3}\text{m}}\right)\\ =0.123\text{mm}\end{array}$

Conclusion:

The elongation in a bar when clamped at one end and a stretching force of $980\text{ N}$ is applied at the other end is $0.123\text{ mm}$.