A spring mass system has time period of `2` second. What should be the spring constant of spring if the mass of the block is `10 grams` ?

Explanation

To solve the problem step by step, we will use the formula for the time period of a spring-mass system in simple harmonic motion.

### Step 1: Write down the formula for the time period
The time period \( T \) of a spring-mass system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where:
- \( T \) is the time period,
- \( m \) is the mass of the block,
- \( k \) is the spring constant.

### Step 2: Rearrange the formula to solve for \( k \)
To find the spring constant \( k \), we can rearrange the formula:
\[ k = \frac{4\pi^2 m}{T^2} \]

### Step 3: Convert the mass into kilograms
The mass is given as \( 10 \) grams. We need to convert this into kilograms:
\[ m = 10 \text{ grams} = 0.01 \text{ kg} \]

### Step 4: Substitute the values into the formula
Now we can substitute the values of \( m \) and \( T \) into the rearranged formula. We know \( T = 2 \) seconds:
\[ k = \frac{4\pi^2 (0.01)}{(2)^2} \]

### Step 5: Calculate \( k \)
Now we will calculate \( k \):
\[ k = \frac{4\pi^2 (0.01)}{4} \]
\[ k = \pi^2 (0.01) \]
Using the approximate value of \( \pi \approx 3.14 \):
\[ k \approx (3.14)^2 \times 0.01 \]
\[ k \approx 9.86 \times 0.01 \]
\[ k \approx 0.0986 \text{ N/m} \]

### Step 6: Round off the answer
Rounding off \( 0.0986 \) N/m gives us:
\[ k \approx 0.1 \text{ N/m} \]

Thus, the spring constant \( k \) is approximately \( 0.1 \) N/m.

### Final Answer:
The spring constant of the spring should be \( 0.1 \text{ N/m} \).

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