Question

Two springs of equal length and equal area of cross-section are suspended from a rigid support and loaded with same mass. Young’s modulus of two springs is in the ratio $4:3$. When the springs are stretched through some distance and released, they will oscillate with periods $T_1$ and $T_2$. The ratio $T_1 : T_2$ is

• A. $2 : \sqrt{3}$
• B. $3 : 4$
• C. $4 : 3$
• D. $\sqrt{3} : 2$

Hint:

For spring oscillations, the time period is inversely proportional to the square root of Young’s modulus.

Answer 1

The Correct Option is D

Step 1: Relation between time period and Young’s modulus.
For a spring-mass system, the time period of oscillation is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \] The spring constant $k$ for a wire is proportional to Young’s modulus $Y$.
Step 2: Express time period in terms of Young’s modulus.
Since $T \propto \frac{1}{\sqrt{k}}$ and $k \propto Y$, we get: \[ T \propto \frac{1}{\sqrt{Y}} \] Step 3: Use the given ratio of Young’s moduli.
\[ Y_1 : Y_2 = 4 : 3 \] \[ T_1 : T_2 = \sqrt{\frac{Y_2}{Y_1}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] Step 4: Final ratio.
\[ T_1 : T_2 = \sqrt{3} : 2 \]