A block of mass 500 g is connected to a spring of spring constant k  312.5 N / m on a frictionless table. The spring is held firmly at the other end. The block is pulled a distance of 5 cm and then released to make SHM. Calculate the time period of its oscillations

Options

A. 2.0 s
B. 1.75 s
C. 0.5 s
D. 0.25 s

Explanation

To solve the problem of calculating the time period of oscillations for a block connected to a spring, we can follow these steps:

### Step 1: Identify the given values
- Mass of the block (m) = 500 g = 0.5 kg (convert grams to kilograms)
- Spring constant (k) = 312.5 N/m
- Displacement (x) = 5 cm (not needed for time period calculation)

### Step 2: Use the formula for the time period of a mass-spring system
The formula for the time period (T) of a mass-spring system undergoing simple harmonic motion is given by:

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

### Step 3: Substitute the values into the formula
Now, substituting the values of mass (m) and spring constant (k) into the formula:

\[ T = 2\pi \sqrt{\frac{0.5 \, \text{kg}}{312.5 \, \text{N/m}}} \]

### Step 4: Calculate the value inside the square root
First, calculate the fraction:

\[ \frac{0.5}{312.5} = 0.0016 \]

### Step 5: Calculate the square root
Now, take the square root of 0.0016:

\[ \sqrt{0.0016} = 0.04 \]

### Step 6: Multiply by \(2\pi\)
Now, multiply by \(2\pi\):

\[ T = 2\pi \times 0.04 \approx 0.2513 \, \text{seconds} \]

### Step 7: Round the answer
Rounding to two decimal places, we find:

\[ T \approx 0.25 \, \text{seconds} \]

### Final Answer
The time period of the oscillations is approximately **0.25 seconds**.

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