Wave Motion 2 Question 12
12. A sonometer wire of length $1.5 m$ is made of steel. The tension in it produces an elastic strain of $1 %$. What is the fundamental frequency of steel if density and elasticity of steel are $7.7 \times 10^{3} kg / m^{3}$ and $2.2 \times 10^{11} N / m^{2}$ respectively?
(a) $188.5 Hz$
(b) $178.2 Hz$
(c) $200.5 Hz$
(d) $770 Hz$
(2013 Main)
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Answer:
Correct Answer: 12. (b)
Solution:
- Fundamental frequency of sonometer wire
$$ f=\frac{v}{2 l}=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}=\frac{1}{2 l} \sqrt{\frac{T}{A d}} $$
Here, $\mu=$ mass per unit length of wire.
Also, Young’s modulus of elasticity $Y=\frac{T l}{A \Delta l}$
$$ \begin{aligned} \Rightarrow \frac{T}{A} & =\frac{Y \Delta l}{l} \Rightarrow f=\frac{1}{2 l} \sqrt{\frac{Y \Delta l}{l d}} \Rightarrow l=1.5 m, \frac{\Delta l}{l}=0.01 \\ d & =7.7 \times 10^{3} kg / m^{3} \Rightarrow Y=2.2 \times 10^{11} N / m^{2} \end{aligned} $$
After substituting the values we get,
$$ f \approx 178.2 Hz $$
