A vertical spring-mass system with lower end of spring is fixed, made to undergo small oscillations. If the spring is stretched by `25cm`, energy stored in the spring is `5J`. Find the mass of the block if it makes `5` oscillations each second.

Explanation

To solve the problem step by step, we will follow the outlined process:

### Step 1: Understand the problem
We have a vertical spring-mass system where the spring is stretched by 25 cm (0.25 m) and the energy stored in the spring is 5 J. We need to find the mass of the block that makes 5 oscillations per second.

### Step 2: Convert the stretch into meters
The stretch in the spring is given as 25 cm. We convert this into meters:
\[ x = 25 \, \text{cm} = 0.25 \, \text{m} \]

### Step 3: Use the formula for potential energy stored in the spring
The potential energy (PE) stored in a spring is given by the formula:
\[ PE = \frac{1}{2} k x^2 \]
Where \( k \) is the spring constant and \( x \) is the stretch in the spring. We know the potential energy is 5 J, so we can set up the equation:
\[ 5 = \frac{1}{2} k (0.25)^2 \]

### Step 4: Solve for the spring constant \( k \)
Rearranging the equation to solve for \( k \):
\[ 5 = \frac{1}{2} k (0.0625) \]
\[ 5 = 0.03125 k \]
\[ k = \frac{5}{0.03125} = 160 \, \text{N/m} \]

### Step 5: Use the frequency to find the mass
The frequency \( f \) of the oscillation is given as 5 Hz. The relationship between frequency and mass is given by:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
Rearranging this for mass \( m \):
\[ m = \frac{k}{(2\pi f)^2} \]

### Step 6: Substitute the values into the equation
Substituting \( k = 160 \, \text{N/m} \) and \( f = 5 \, \text{Hz} \):
\[ m = \frac{160}{(2\pi \cdot 5)^2} \]
Calculating \( 2\pi \cdot 5 \):
\[ 2\pi \cdot 5 = 10\pi \]
Now squaring it:
\[ (10\pi)^2 = 100\pi^2 \]
Thus, substituting back:
\[ m = \frac{160}{100\pi^2} \]
\[ m = \frac{8}{5\pi^2} \]

### Step 7: Final result
The mass of the block is:
\[ m \approx \frac{8}{5\pi^2} \, \text{kg} \]