A body of mass `m` is atteched to the lower end of a spring whose upper end is fixed .The spring has negaligible mass .When the mass `m` is slightly puylled down and released it oscillation with a time period of `3 s` when the mass `m` is increased by `1kg` time period of oscillations becomes `5s` The value of `m` in `kg` is

Options

A. `(3)/(4)`
B. `(4)/(3)`
C. `(16)/(3)`
D. `(9)/(16)`

Explanation

To solve the problem step by step, we will use the formula for the time period of a mass-spring system and set up equations based on the information given.

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

### Step 2: Set up the first equation using the initial mass \( m \).
According to the problem, when the mass is \( m \), the time period is \( 3 \) seconds. Thus, we can write:
\[ 3 = 2\pi \sqrt{\frac{m}{k}} \]

### Step 3: Rearrange the first equation to express \( \frac{m}{k} \).
Squaring both sides gives:
\[ 9 = 4\pi^2 \frac{m}{k} \]
Rearranging this, we find:
\[ \frac{m}{k} = \frac{9}{4\pi^2} \quad \text{(Equation 1)} \]

### Step 4: Set up the second equation using the increased mass \( m + 1 \).
When the mass is increased by \( 1 \) kg, the time period becomes \( 5 \) seconds. Thus, we can write:
\[ 5 = 2\pi \sqrt{\frac{m + 1}{k}} \]

### Step 5: Rearrange the second equation to express \( \frac{m + 1}{k} \).
Squaring both sides gives:
\[ 25 = 4\pi^2 \frac{m + 1}{k} \]
Rearranging this, we find:
\[ \frac{m + 1}{k} = \frac{25}{4\pi^2} \quad \text{(Equation 2)} \]

### Step 6: Set up a ratio of the two equations.
Now, we can set up a ratio of Equation 1 and Equation 2:
\[ \frac{\frac{m}{k}}{\frac{m + 1}{k}} = \frac{9/4\pi^2}{25/4\pi^2} \]
This simplifies to:
\[ \frac{m}{m + 1} = \frac{9}{25} \]

### Step 7: Cross-multiply to solve for \( m \).
Cross-multiplying gives:
\[ 25m = 9(m + 1) \]
Expanding this results in:
\[ 25m = 9m + 9 \]
Subtracting \( 9m \) from both sides gives:
\[ 16m = 9 \]

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

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

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