Question

A metallic wire with tension T and at temperature $30^\circ {\text{C}}$ vibrates with its fundamental frequency of 1 KHz. The same wire with the same tension but at $10^\circ {\text{C}}$ temperature vibrates with a fundamental frequency of 1.001 KHz. The coefficient of linear expansion of the wire is:
A. $2 \times {10^{ - 4}}/^\circ {\text{C}}$
B. $1.5 \times {10^{ - 4}}/^\circ {\text{C}}$
C. $1 \times {10^{ - 4}}/^\circ {\text{C}}$
D. $0.5 \times {10^{ - 4}}/^\circ {\text{C}}$

Answer 1

Use the expression for frequency of vibrations to express the fundamental frequencies at the respective temperatures and take the ratio. Use the formula for linear expansion of the material to express the ratio of the final length to the initial length of the metallic wire. Consider $30^\circ {\text{C}}$ as the initial temperature and $10^\circ {\text{C}}$ as the final temperature.

Formula used:
The frequency of the vibration is given as,
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{l}}$
Here, g is the acceleration due to gravity and l is the length of the wire.
$l = {l_0}\left( {1 + \alpha \,\Delta T} \right)$
Here, l is the final length, ${l_0}$ is the initial length, $\alpha$ is the linear expansion coefficient and $\Delta T$ is the change in temperature.

Complete step by step answer:
We have given the initial temperature of the wire is ${T_1} = 30^\circ {\text{C}}$ and the final temperature of the wire is ${T_2} = 10^\circ {\text{C}}$. At ${T_1} = 30^\circ {\text{C}}$, the wire vibrates at the fundamental frequency${f_1} = 1\,{\text{KHz}}$ and at ${T_2} = 10^\circ {\text{C}}$, the wire vibrates at the fundamental frequency ${f_2} = 1.001\,{\text{KHz}}$.

We know the expression for the frequency of the vibration is given as,
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{l}}$
Here, g is the acceleration due to gravity and l is the length of the wire.

Let’s express the fundamental frequency of the vibration at ${T_1} = 30^\circ {\text{C}}$ as follows,
${f_1} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_1}}}}$ …… (1)

Let’s express the fundamental frequency of the vibration at ${T_2} = 10^\circ {\text{C}}$ as follows,
${f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_2}}}}$ …… (2)
Dividing equation (2) by equation (1), we get,
$\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_1}}}} }}{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_2}}}} }}$
$\Rightarrow \dfrac{{{f_1}}}{{{f_2}}} = \sqrt {\dfrac{{{l_2}}}{{{l_1}}}}$
$\Rightarrow \dfrac{{{l_2}}}{{{l_1}}} = {\left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)^2}$ …… (3)

We can express the linear expansion of the wire due to change in temperature as,
${l_2} = {l_1}\left( {1 + \alpha \,\Delta T} \right)$
$\Rightarrow \dfrac{{{l_2}}}{{{l_1}}} = \left( {1 + \alpha \,\Delta T} \right)$
Here, $\alpha$ is the linear expansion coefficient and $\Delta T$ is the change in temperature.

Using equation (3), we can rewrite the above equation as,
$\dfrac{{f_1^2}}{{f_2^2}} = 1 + \alpha \,\Delta T$

Substituting 1 KHz for ${f_1}$, 1.001 KHz for ${f_2}$ and $- 20^\circ {\text{C}}$ for $\Delta T$ in the above equation, we get,
$\dfrac{1}{{{{\left( {1.001} \right)}^2}}} = 1 + \alpha \left( { - 20} \right)$
$\Rightarrow 0.998 = 1 - 20\alpha$
$\Rightarrow \alpha = \dfrac{{2 \times {{10}^{ - 3}}}}{{20}}$
$\therefore \alpha = 1 \times {10^{ - 4}}/^\circ {\text{C}}$

So, the correct answer is option C.

Note: We have assumed the final temperature as $10^\circ {\text{C}}$ and the initial temperature as $30^\circ {\text{C}}$. Therefore, the change in temperature is $- 20^\circ {\text{C}}$. You don’t need to convert the temperature in Kelvin since the coefficient of linear expansion is in $/^\circ C$.