Question

A wire of density \[9 \times {10^3}kg{m^{ - 3}}\] is stretched between two clamps 1m apart. The resulting strain in the wire is \[4.9 \times {10^{ - 4}}\]. Find the lowest frequency of the transverse vibrations in the wire (Young’s modulus of wire \[Y = 9 \times {10^{10}}N{m^2}\]), (to the nearest integer),____________.

Answer 1

Hint:Stress is defined as force applied on the material per unit area of cross-section. Strain is the ratio of change in length to its original length. Strain is dimensionless where the S.I. the unit of stress is \[N{m^2}\]. As we know, the frequency of the wave is the number of oscillations per second.Here we are going to use the expression of the frequency of a wave.

Formula Used:
To find the frequency of a wave we have,
\[f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{\mu }} \]
Where, T is the tension in the wire, L is length of wire and \[\mu \] is mass per unit length.

Complete step by step solution:
Consider a wire which has been stretched between the two clamps 1m apart having a density of \[9 \times {10^3}kg{m^{ - 3}}\]. Then the resulting strain in the wire is \[4.9 \times {10^{ - 4}}\]. We need to find the lowest frequency of the transverse vibrations in the wire. In order to find the frequency, first, we need to find the mass per unit length of the wire and tension.

So, the mass per unit length is,
\[\mu = \rho A\]
And, \[Y = \left( {\dfrac{{\dfrac{T}{A}}}{{\dfrac{{\Delta l}}{l}}}} \right)\] that is young’s modulus is the ratio of stress to strain.
\[T = \dfrac{{Y\Delta lA}}{l}\]
Now, the frequency of a wave is,
\[f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{\mu }} \]…… (1)
Now, substitute the value of \[\mu \]and T in equation (1) we get,
\[f = \dfrac{1}{{2l}}\sqrt {\dfrac{{\dfrac{{Y\Delta lA}}{l}}}{{\rho A}}} \]
\[\Rightarrow f = \dfrac{1}{{2l}}\sqrt {\dfrac{{Y\Delta l}}{{\rho l}}} \]

Substitute the value of \[Y = 9 \times {10^{10}}N{m^2}\], \[\Delta l = 4.9 \times {10^{ - 4}}\], \[\rho = 9 \times {10^3}kg{m^{ - 3}}\] and \[l = 1m\] we get,
\[\Rightarrow f = \dfrac{1}{{2 \times 1}}\sqrt {\dfrac{{9 \times {{10}^{10}} \times 4.9 \times {{10}^{ - 4}}}}{{9 \times {{10}^3} \times 1}}} \]
\[\Rightarrow f = \dfrac{1}{{2 \times 1}}\sqrt {4.9 \times {{10}^{ - 3}}} \]
\[\Rightarrow f = \dfrac{1}{{2 \times 1}} \times 70\]
\[\therefore f = 35Hz\]

Therefore, the lowest frequency of the transverse vibrations in the wire is 35 Hz.

Note:In order to resolve this problem, it is important to know about the formula for the frequency of a wave, tension in the wire and as well as for young’s modulus. By knowing all these formulas, we can easily solve the problem.