Question

A uniform circular disc of mass $12 \text{ kg}$ is held by two identical springs. When the disc is slightly pressed down and released, it executes S.H.M. of period $2 \text{ second}$. The force constant of each spring is (Take $\pi^2 = 10$)}

• A. $230 \text{ Nm}^{-1}$
• B. $120 \text{ Nm}^{-1}$
• C. $60 \text{ Nm}^{-1}$
• D. $30 \text{ Nm}^{-1}$

Hint:

For springs in parallel, the effective spring constant is the sum of individual spring constants. The period of oscillation for a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k_{\text{eff$. Ensure correct identification of $k_{\text{eff$ for the system.

Answer 1

The Correct Option is A

Concept:
For a mass attached to springs performing S.H.M., time period is: \[ T=2\pi\sqrt{\frac{m}{k_{\text{eff \] where:

• $T$ = time period
• $m$ = mass attached
• $k_{\text{eff$ = effective spring constant

For two identical springs connected in parallel: \[ k_{\text{eff=k+k=2k \]
Step 1: Find effective spring constant
Each spring has spring constant $k$. Since springs are parallel: \[ k_{\text{eff=2k \]
Step 2: Use time period formula
Given: \[ m=12\,kg,\qquad T=2\,s \] Substitute in formula: \[ 2=2\pi\sqrt{\frac{12}{2k \] Divide both sides by $2\pi$: \[ \frac{1}{\pi}=\sqrt{\frac{12}{2k \]
Step 3: Square both sides
\[ \frac{1}{\pi^2}=\frac{12}{2k} \] \[ \frac{1}{\pi^2}=\frac{6}{k} \] \[ k=6\pi^2 \]
Step 4: Use } $\pi^2=10$}
\[ k=6\times 10 \] \[ k=60\,N\,m^{-1} \]
Step 5: Final Answer
Spring constant of each spring is: \[ \boxed{60\ N\,m^{-1 \] Quick Tip:
Parallel springs become stiffer, so their spring constants add directly.