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. `(3)/(5)`
B. `(24)/(9)`
C. `(16)/(9)`
D. `(5)/(3)`

Explanation

To solve the problem step by step, we will use the formulas for the time period of a mass-spring system and manipulate them to find the required ratio of \( \frac{m}{M} \).

### Step 1: Write the formula for the time period of the mass-spring system.
The time period \( T \) of a mass \( M \) attached to a spring with spring constant \( k \) is given by:
\[ T = 2\pi \sqrt{\frac{M}{k}} \]

### Step 2: Write the time period for the initial mass \( M \).
From the problem, we know that the initial time period is \( T \):
\[ T = 2\pi \sqrt{\frac{M}{k}} \quad \text{(Equation 1)} \]

### Step 3: Write the time period for the new mass \( M + m \).
When the mass is increased by \( m \), the new time period becomes \( \frac{5T}{3} \):
\[ \frac{5T}{3} = 2\pi \sqrt{\frac{M + m}{k}} \quad \text{(Equation 2)} \]

### Step 4: Divide Equation 2 by Equation 1.
We will divide the two equations to eliminate \( k \) and \( 2\pi \):
\[ \frac{\frac{5T}{3}}{T} = \frac{2\pi \sqrt{M + m}}{2\pi \sqrt{M}} \]
This simplifies to:
\[ \frac{5}{3} = \frac{\sqrt{M + m}}{\sqrt{M}} \]

### Step 5: Square both sides to eliminate the square root.
Squaring both sides gives:
\[ \left(\frac{5}{3}\right)^2 = \frac{M + m}{M} \]
This simplifies to:
\[ \frac{25}{9} = \frac{M + m}{M} \]

### Step 6: Cross-multiply to solve for \( m \).
Cross-multiplying gives:
\[ 25M = 9(M + m) \]
Expanding the right side:
\[ 25M = 9M + 9m \]

### Step 7: Rearrange the equation to isolate \( m \).
Rearranging gives:
\[ 25M - 9M = 9m \]
This simplifies to:
\[ 16M = 9m \]

### Step 8: Solve for the ratio \( \frac{m}{M} \).
Dividing both sides by \( M \) gives:
\[ \frac{m}{M} = \frac{16}{9} \]

### Final Answer:
The ratio \( \frac{m}{M} \) is:
\[ \frac{m}{M} = \frac{16}{9} \]