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 .

Options

A. (a) `3/5`
B. (b) (25)/9`
C. (c ) (16)/9`
D. (d) `5/3`

Explanation

To solve the problem, we need to analyze the relationship between the mass, the spring constant, and the time period of the simple harmonic motion (SHM).

### Step-by-step Solution:

1. **Understanding the Time Period Formula**:
The time period \( T \) of a mass \( M \) attached to a spring with spring constant \( k \) is given by the formula:
\[ T = 2\pi \sqrt{\frac{M}{k}} \]

2. **Initial Time Period**:
For the initial mass \( M \), the time period is:
\[ T = 2\pi \sqrt{\frac{M}{k}} \]

3. **Increased Mass**:
When the mass is increased by \( m \), the new total mass becomes \( M + m \). The new time period \( T' \) is given as:
\[ T' = \frac{5T}{3} \]

4. **New Time Period Formula**:
The new time period can also be expressed using the new mass:
\[ T' = 2\pi \sqrt{\frac{M + m}{k}} \]

5. **Setting Up the Equation**:
We can set the two expressions for the new time period equal to each other:
\[ \frac{5T}{3} = 2\pi \sqrt{\frac{M + m}{k}} \]

6. **Substituting for \( T \)**:
Substitute \( T = 2\pi \sqrt{\frac{M}{k}} \) into the equation:
\[ \frac{5}{3} \cdot 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M + m}{k}} \]

7. **Canceling Common Terms**:
We can cancel \( 2\pi \) and \( \sqrt{k} \) from both sides:
\[ \frac{5}{3} \sqrt{M} = \sqrt{M + m} \]

8. **Squaring Both Sides**:
To eliminate the square roots, we square both sides:
\[ \left(\frac{5}{3}\right)^2 M = M + m \]
\[ \frac{25}{9} M = M + m \]

9. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{25}{9} M - M = m \]
\[ \frac{25}{9} M - \frac{9}{9} M = m \]
\[ \frac{16}{9} M = m \]

10. **Finding the Ratio \( \frac{m}{M} \)**:
Now, we can find the ratio \( \frac{m}{M} \):
\[ \frac{m}{M} = \frac{16}{9} \]

### Final Answer:
The ratio of \( \frac{m}{M} \) is \( \frac{16}{9} \).

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