Simple Harmonic Motion Ques 22
- 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)
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Answer:
Correct Answer: 22.$f=\frac{1}{2 \pi} \sqrt{\frac{Y A}{m L}}$
Solution:
Formula:
- Force constant of a wire $k=\frac{Y A}{L}$
Frequency of oscillation $f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$
or $\quad f=\frac{1}{2 \pi} \sqrt{\frac{(Y A / L)}{m}}=\frac{1}{2 \pi} \sqrt{\frac{Y A}{m L}}$
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