A light spring of force constant `8 Nm^(-1)` is cut into two equal halves and the two are connected in parallel, the equivalent force constant of the system is

Options

A. `16 Nm^(-1)`
B. `32 Nm^(-1)`
C. `8 Nm^(-1)`
D. `24 Nm^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a spring with a force constant (spring constant) \( k = 8 \, \text{N/m} \). This spring is cut into two equal halves, and we need to find the equivalent spring constant when these two halves are connected in parallel.

### Step 2: Determine the new spring constant after cutting
When a spring is cut into two equal halves, the new spring constant \( k' \) of each half can be calculated using the relationship between the spring constant and the length of the spring. The formula is:
\[ k \cdot L = k' \cdot \frac{L}{2} \]
Where:
- \( k \) is the original spring constant (8 N/m)
- \( L \) is the original length of the spring
- \( k' \) is the new spring constant of each half
- \( \frac{L}{2} \) is the length of each half

Rearranging the formula gives:
\[ k' = \frac{2k}{1} = 2k \]
Substituting the value of \( k \):
\[ k' = 2 \times 8 = 16 \, \text{N/m} \]

### Step 3: Calculate the equivalent spring constant in parallel
When two springs are connected in parallel, the equivalent spring constant \( k_{eq} \) is simply the sum of the individual spring constants:
\[ k_{eq} = k_1 + k_2 \]
Since both halves have the same spring constant \( k' = 16 \, \text{N/m} \):
\[ k_{eq} = 16 + 16 = 32 \, \text{N/m} \]

### Final Answer
The equivalent force constant of the system is \( 32 \, \text{N/m} \).

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