Rankers Physics
Topic: Oscillation
Subtopic: Spring Block System

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 SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. Then the ratio of m/M is :
3/5
25/9
16/9
5/3

Solution:

1. Initial Time Period:
\[
T = 2\pi \sqrt{\frac{M}{k}}
\]

2. New Time Period:
\[
\frac{5T}{3} = 2\pi \sqrt{\frac{M + m}{k}}
\]

3. Divide the New Time Period by the Initial Time Period:

\[
\frac{\frac{5T}{3}}{T} = \sqrt{\frac{M + m}{M}}
\]

\[
\frac{5}{3} = \sqrt{\frac{M + m}{M}}
\]

4. Square Both Sides:

\[
\frac{25}{9} = \frac{M + m}{M}
\]

5. Solve for \( \frac{m}{M} \):

\[
\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}
\]

Answer:

\[
\frac{m}{M} = \frac{16}{9}
\]

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