Question

Three thin wires of equal length in suspension have the respective ratio of their area of cross section \(1:2:4\) and Young's moduli \(4:2:1\), then the ratio of their time periods due to attached at the other ends due to elongation is

• A. \(1:1:1\)
• B. \(1:2:4\)
• C. \(4:2:1\)
• D. \(2:\sqrt{2}:1\)
• E. \(1:\sqrt{2}:2\)

Hint:

For wire oscillation, compare the product \(AY\). If \(AY\) is the same, the time period is the same.

Answer 1

The Correct Option is A

For oscillation of a mass suspended by a wire: \[ T=2\pi\sqrt{\frac{mL}{AY}} \] Since the length \(L\) and attached mass \(m\) are the same, \[ T\propto \frac{1}{\sqrt{AY}} \] Now compute \(AY\) for the three wires: First wire: \[ A_1Y_1=1\times 4=4 \] Second wire: \[ A_2Y_2=2\times 2=4 \] Third wire: \[ A_3Y_3=4\times 1=4 \] So all have the same value of \(AY\), hence all have the same time period. Therefore, \[ \boxed{(A)\ 1:1:1} \]