A body of mass 4.9 kg hangs from a spring and oscillates with a period 0.5 s. On the removal of the body, the spring is shortened by (Take `g=10m s^(-2), pi^(2)=10`)

Options

A. 0.625 cm
B. 0.63 cm
C. 6.25 cm
D. 63 cm

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the spring constant (k)
The formula for the period \( T \) of a mass-spring system is given by:

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

Where:
- \( T \) is the period (0.5 s),
- \( m \) is the mass (4.9 kg),
- \( k \) is the spring constant.

Rearranging this formula to solve for \( k \):

\[ k = \frac{4\pi^2 m}{T^2} \]

### Step 2: Substitute the values into the equation
Substituting the known values into the equation:

\[ k = \frac{4 \times 10 \times 4.9}{(0.5)^2} \]

Calculating \( (0.5)^2 \):

\[ (0.5)^2 = 0.25 \]

Now substituting this back into the equation for \( k \):

\[ k = \frac{4 \times 10 \times 4.9}{0.25} \]

### Step 3: Calculate \( k \)
Calculating the numerator:

\[ 4 \times 10 \times 4.9 = 196 \]

Now divide by 0.25:

\[ k = \frac{196}{0.25} = 784 \, \text{N/m} \]

### Step 4: Calculate the extension of the spring when the mass is hanging
The force due to the weight of the mass is given by:

\[ F = mg \]

Substituting the values:

\[ F = 4.9 \times 10 = 49 \, \text{N} \]

Using Hooke's Law \( F = kx \) to find the extension \( x \):

\[ x = \frac{F}{k} = \frac{49}{784} \]

### Step 5: Calculate the extension \( x \)
Now, calculating \( x \):

\[ x = \frac{49}{784} = 0.0625 \, \text{m} = 6.25 \, \text{cm} \]

### Step 6: Determine the shortening of the spring upon removal of the mass
When the mass is removed, the spring will return to its original length, which means it shortens by the amount it was extended, which is \( 6.25 \, \text{cm} \).

### Final Answer
The spring is shortened by **6.25 cm** upon the removal of the body.

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