A 5 kg collar is attached to a spring of spring constant 500 N `m^(-1)`. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10 cm and released. The time period of oscillation is

Options

A. `(pi)/(5)s`
B. `(pi)/(2)s`
C. `pis`
D. `2pis`

Explanation

To find the time period of oscillation for the collar attached to the spring, we can use the formula for the time period \( T \) of a mass-spring system:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

where:
- \( T \) is the time period,
- \( m \) is the mass of the collar,
- \( k \) is the spring constant.

### Step 1: Identify the given values
- Mass of the collar, \( m = 5 \, \text{kg} \)
- Spring constant, \( k = 500 \, \text{N/m} \)

### Step 2: Substitute the values into the formula
Now, substituting the values of \( m \) and \( k \) into the formula:

\[ T = 2\pi \sqrt{\frac{5 \, \text{kg}}{500 \, \text{N/m}}} \]

### Step 3: Simplify the fraction inside the square root
Calculating the fraction:

\[ \frac{5}{500} = \frac{1}{100} \]

### Step 4: Take the square root
Now, taking the square root:

\[ \sqrt{\frac{1}{100}} = \frac{1}{10} \]

### Step 5: Multiply by \( 2\pi \)
Now, substitute back into the equation for \( T \):

\[ T = 2\pi \cdot \frac{1}{10} = \frac{2\pi}{10} = \frac{\pi}{5} \]

### Step 6: Conclusion
Thus, the time period of oscillation is:

\[ T = \frac{\pi}{5} \, \text{seconds} \]

### Final Answer
The time period of oscillation is \( \frac{\pi}{5} \) seconds.

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