A 4-kg block extends a spring by 16 cm from its unstretched postion .The block is removed and a 0.5 -kg boby is hung from the same spring If the spring is then stretched and released , what is the time period of oscillations of the mass suspendrd ? What is the force constant of the spring ?

Answer

To solve the problem step by step, we will first determine the force constant of the spring and then calculate the time period of oscillation for the 0.5 kg mass.

### Step 1: Calculate the force constant (k) of the spring

Given:
- Mass (m) = 4 kg
- Extension (x) = 16 cm = 0.16 m
- Gravitational acceleration (g) = 9.81 m/s² (approximately)

Using Hooke's Law, the force exerted by the spring is equal to the weight of the block at equilibrium:
\[ kx = mg \]

Rearranging the formula to find k:
\[ k = \frac{mg}{x} \]

Substituting the known values:
\[ k = \frac{4 \, \text{kg} \times 9.81 \, \text{m/s}^2}{0.16 \, \text{m}} \]
\[ k = \frac{39.24 \, \text{N}}{0.16 \, \text{m}} \]
\[ k = 245.25 \, \text{N/m} \]

### Step 2: Calculate the time period (T) of oscillation for the 0.5 kg mass

Given:
- New mass (m') = 0.5 kg

The formula for the time period of a mass-spring system is:
\[ T = 2\pi \sqrt{\frac{m'}{k}} \]

Substituting the known values:
\[ T = 2\pi \sqrt{\frac{0.5 \, \text{kg}}{245.25 \, \text{N/m}}} \]

Calculating the square root:
\[ T = 2\pi \sqrt{0.00204} \]
\[ T = 2\pi \times 0.0452 \]
\[ T \approx 0.283 \, \text{s} \]

### Final Answers:
- Force constant of the spring (k) = 245.25 N/m
- Time period of oscillation (T) = 0.283 s

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