Simple Harmonic Motion 3 Question 14
14. A point mass $m$ is suspended at the end of massless wire of length $L$ and cross-sectional area $A$. If $Y$ is the Young’s modulus of elasticity of the material of the wire, obtain the expression for the frequency of the simple harmonic motion along the vertical line.
(1978)
Integer Answer Type Question
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Answer:
Correct Answer: 14. $f=\frac{1}{2 \pi} \sqrt{\frac{Y A}{m L}}$
Solution:
- Force constant of a wire $k=\frac{Y A}{L}$
Frequency of oscillation $f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$
or $\quad f=\frac{1}{2 \pi} \sqrt{\frac{(Y A / L)}{m}}=\frac{1}{2 \pi} \sqrt{\frac{Y A}{m L}}$
