Question

A small mass \( m \) is suspended at the end of a wire having negligible mass, length \( L \) and cross-sectional area \( A \). The frequency of oscillation for the S.H.M. along the vertical line is ( \( Y \) = Young’s modulus of the wire )

• A. \( \dfrac{1}{2\pi}\left(\dfrac{YA}{mL}\right)^{1/2} \)
• B. \( \dfrac{2\pi YA}{mL} \)
• C. \( \dfrac{YA}{2\pi mL} \)
• D. \( 2\pi \left(\dfrac{YA}{mL}\right)^{1/2} \)

Hint:

A stretched wire behaves like a spring with spring constant \( k = \dfrac{YA}{L} \).

Answer 1

The Correct Option is A

Step 1: Find the restoring force due to extension of wire.
When the mass is displaced vertically, the wire is stretched. The restoring force due to elasticity is \[ F = \frac{YA}{L}\,x \] where \( x \) is the extension.
Step 2: Identify the effective spring constant.
Comparing with Hooke’s law \( F = kx \), we get \[ k = \frac{YA}{L} \]
Step 3: Write frequency formula for SHM.
For a mass–spring system, \[ f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \]
Step 4: Substitute the value of \( k \).
\[ f = \frac{1}{2\pi}\sqrt{\frac{YA}{mL}} \]
Step 5: Conclusion.
The frequency of oscillation is \[ f = \frac{1}{2\pi}\left(\frac{YA}{mL}\right)^{1/2}. \]