Question

A wire of density $9 \times {10^3}kg/{m^3}$stretched between two clamps $1m$ apart is subjected to an extension of $4.9 \times {10^{ - 4}}m$. If Young's modulus of the wire is $9 \times {10^{10}}m$. The lowest frequency of transverse vibrations in the wire is
A. $35\,Hz$
B. $70\,Hz$
C. $105\,Hz$
D. $140\,Hz$

Answer 1

Hint:In the case of a problem based on mechanical properties of solid, we know that Young’s modulus plays a significant role in establishing a relationship between different parameters hence, identify the formula of Young’s modulus that should be used to calculate the frequency of transverse vibration in a wire in order to provide the accurate solution.

Formula used:
The frequency of transverse vibration is given as,
$f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $
Here, $T$ is the tension in the string, $m$ is the mass per unit length and $l$ is the length of the string.

Complete step by step solution:
Wire Density $\rho = 9 \times {10^3}kg/{m^3}$........(given)
A Wire is subjected to an extension i.e., a change in length of the wire is,
$\Delta l = 4.9 \times {10^{ - 4}}m$..........(given)
Young’s modulus of wire $Y = 9 \times {10^{10}}m$......(given)
As, wire is stretched between two clamps i.e., $1m$ apart, therefore, length should be considered as $l = 1m$...(given)
We know that mass per unit length
$m = \dfrac{{\text{Volume} \times \text{Density}}}{\text{length}} = \dfrac{{A \times l \times \rho }}{l} = A\rho $… (1)
where, A = Area of the cross-section of a wire.

Now, the formula to calculate the frequency of transverse vibration is given as:
$f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $ … (2)
Also, we know that the expression for Young’s modulus can be given as:
$Y = \dfrac{\dfrac{T}{A}}{\dfrac{\Delta l}{l}}$
$\Rightarrow T = \dfrac{{YA\Delta l}}{l}$

Substitute the values of $m$and $T$ in eq. (2), we get
$f = \dfrac{1}{{2l}}\sqrt {\dfrac{{\dfrac{{YA\Delta l}}{l}}}{{A\rho }}} \\ $
$\Rightarrow f = \dfrac{1}{{2l}}\sqrt {\dfrac{{Y\Delta l}}{{l\rho }}} \\ $
Substitute all the given values in the question in above expression, we get
$\Rightarrow f = \dfrac{1}{{2(1)}}\sqrt {\dfrac{{9 \times {{10}^{10}} \times 4.9 \times {{10}^{ - 4}}}}{{1 \times 9 \times {{10}^3}}}} \\ $
$\Rightarrow f = \dfrac{1}{2}\sqrt {\dfrac{{9 \times 49 \times {{10}^2}}}{9}}\\ $
$\Rightarrow f = \dfrac{{7 \times 10}}{2} \\$
$\therefore f = 35\,Hz$
Thus, the lowest frequency of transverse vibrations in the wire is $35\,Hz$.

Hence, the correct option is A.

Note: Since this is a problem related to transverse vibration in mechanics, given conditions are to be analyzed very carefully and quantities required to calculate the lowest frequency must be identified on a prior basis as it gives a better understanding of the problem. Units must be put after each end result.