Question

The time period of SHM is \( 2 \, \mathrm{s} \) with mass \( m \). If an additional mass of \( 40 \, \mathrm{g} \) is added, the time period increases by \( 3 \, \mathrm{s} \). What is \( m \) (in grams)?

• A. \( 7.64 \, \mathrm{g} \)
• B. \( 40 \, \mathrm{g} \)
• C. \( 50 \, \mathrm{g} \)
• D. \( 60 \, \mathrm{g} \)

Hint:

N/A

Answer 1

The Correct Option is A

The time period of a simple harmonic motion (SHM) is given by: \[ T = 2\pi \sqrt{\frac{m}{k}}, \] where: \( T \) is the time period, \( m \) is the mass, \( k \) is the spring constant.

 Step 1: Initial time period. 
The time period with mass \( m \) is: \[ T_1 = 2 = 2\pi \sqrt{\frac{m}{k}}. \] Squaring both sides: \[ T_1^2 = \frac{4\pi^2 m}{k}. \] Rearranging for \( m \): \[ m = \frac{k T_1^2}{4\pi^2}. \] 

Step 2: New time period with added mass. 
The new time period is \( T_2 = 5 \, \mathrm{s} \), and the total mass becomes \( m + 0.04 \): \[ T_2^2 = \frac{4\pi^2 (m + 0.04)}{k}. \] Rearranging: \[ m + 0.04 = \frac{k T_2^2}{4\pi^2}. \] 

Step 3: Subtract initial from new equation. 
From the two equations: \[ \frac{k T_2^2}{4\pi^2} - \frac{k T_1^2}{4\pi^2} = 0.04. \] Factorize: \[ \frac{k}{4\pi^2} (T_2^2 - T_1^2) = 0.04. \] Substitute \( T_1 = 2 \, \mathrm{s} \) and \( T_2 = 5 \, \mathrm{s} \): \[ \frac{k}{4\pi^2} (5^2 - 2^2) = 0.04. \] Simplify: \[ \frac{k}{4\pi^2} \cdot (25 - 4) = 0.04. \] \[ \frac{k}{4\pi^2} \cdot 21 = 0.04. \] Solve for \( \frac{k}{4\pi^2} \): \[ \frac{k}{4\pi^2} = \frac{0.04}{21}. \] 

Step 4: Substitute back to find \( m \). 
Using the expression for \( m \): \[ m = \frac{k T_1^2}{4\pi^2}. \] Substitute \( T_1 = 2 \, \mathrm{s} \): \[ m = \left(\frac{0.04}{21}\right) \cdot 4. \] Simplify: \[ m = \frac{0.04 \cdot 4}{21} = 0.00761 \, \mathrm{kg}. \] Convert to grams: \[ m = 7.64 \, \mathrm{g}. \] Thus, the mass is \( \mathbf{7.64 \, \mathrm{g}} \).