Question

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

• A. \(\dfrac{1}{2\pi}\left[\dfrac{YAL}{m}\right]^{1/2}\)
• B. \(\dfrac{1}{2\pi}\left[\dfrac{YA}{mL}\right]^{1/2}\)
• C. \(\dfrac{1}{2\pi}\left[\dfrac{mA}{YL}\right]^{1/2}\)
• D. \(\dfrac{1}{2\pi}\left[\dfrac{YL}{mA}\right]^{1/2}\)

Hint:

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

Answer 1

The Correct Option is B

Step 1: Expression for force constant of the wire.
For a stretched wire, the effective force constant is:
\[ k = \frac{YA}{L} \]
Step 2: Time period of vertical SHM.
Time period for a mass–spring system is:
\[ T = 2\pi\sqrt{\frac{m}{k}} \]
Step 3: Substitute the value of \(k\).
\[ T = 2\pi\sqrt{\frac{mL}{YA}} \]
Step 4: Write expression for frequency.
\[ f = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{YA}{mL}} \]
Step 5: Conclusion.
The correct expression for frequency is option (B).