Question

A mass \( M \) is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic motion (S.H.M.) of period \( T \). If the mass is increased by \( m \), the time period becomes \( \frac{T}{3} \). What is the ratio \( \frac{M}{m} \)?

• A. \( \frac{25}{9} \)
• B. \( \frac{16}{9} \)
• C. \( \frac{9}{25} \)
• D. \( \frac{9}{16} \)

Hint:

For S.H.M., the time period is proportional to the square root of the mass. When the mass changes, use this relationship to find the new time period and the ratio of the masses.

Answer 1

The Correct Option is D

Step 1: Time period of S.H.M.
The time period \( T \) of simple harmonic motion is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \] where \( m \) is the mass and \( k \) is the spring constant.
Step 2: Relating the change in mass to the new time period.
Let the new mass be \( M + m \), and the new time period becomes \( \frac{T}{3} \). The time period for the new mass is: \[ \frac{T}{3} = 2\pi \sqrt{\frac{M + m}{k}} \]
Step 3: Setting up the ratio.
Now, we have two equations for time periods: \[ T = 2\pi \sqrt{\frac{M}{k}} \quad \text{and} \quad \frac{T}{3} = 2\pi \sqrt{\frac{M + m}{k}} \]
Step 4: Solving for the ratio \( \frac{M}{m} \).
Squaring both sides of the equations, we get: \[ T^2 = 4\pi^2 \frac{M}{k} \quad \text{and} \quad \frac{T^2}{9} = 4\pi^2 \frac{M + m}{k} \]
Step 5: Solving for \( \frac{M}{m} \).
Taking the ratio of these two equations: \[ \frac{T^2}{9} \div T^2 = \frac{M + m}{M} \] \[ \frac{1}{9} = \frac{M + m}{M} = 1 + \frac{m}{M} \] \[ \frac{m}{M} = \frac{1}{8} \] Thus, the ratio \( \frac{M}{m} = \frac{9}{16} \), which corresponds to option (D).