SATHEE: Simple Harmonic Motion 3 Question 14

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)

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Answer:

Correct Answer: 14. $f = \frac{1}{2 π} \sqrt{\frac{Y A}{m L}}$

Solution:

Force constant of a wire $k = \frac{Y A}{L}$

Frequency of oscillation $f = \frac{1}{2 π} \sqrt{\frac{k}{m}}$

or $f = \frac{1}{2 π} \sqrt{\frac{(Y A / L)}{m}} = \frac{1}{2 π} \sqrt{\frac{Y A}{m L}}$

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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.

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Answer:

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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.