Write the expression for equivalent spring constant of
`(i)` parallel combination of `n-`springs
`(ii)` series combination of `n-`springs

Answer

To find the expressions for the equivalent spring constant for both the parallel and series combinations of n springs, we can break down the solution into two parts.

### Part (i): Parallel Combination of n Springs

1. **Understanding the Setup**:
- When springs are arranged in parallel, they all share the same displacement when a force is applied.
- Let's denote the spring constant of each individual spring as \( k \).

2. **Calculating the Equivalent Spring Constant**:
- In a parallel combination, the total force exerted by the springs is the sum of the forces exerted by each spring.
- According to Hooke's Law, the force exerted by a spring is given by \( F = k \cdot x \), where \( x \) is the displacement.
- If there are \( n \) springs, the total force can be expressed as:
\[ F_{\text{total}} = n \cdot (k \cdot x) = n \cdot k \cdot x \]
- Therefore, the equivalent spring constant \( k_{\text{eq}} \) for the parallel combination is:
\[ k_{\text{eq}} = n \cdot k \]

### Part (ii): Series Combination of n Springs

1. **Understanding the Setup**:
- In a series arrangement, the springs are connected end-to-end, and the same force acts on each spring.
- Again, let the spring constant of each spring be \( k \).

2. **Calculating the Equivalent Spring Constant**:
- For springs in series, the total extension \( x \) is the sum of the extensions of each spring:
\[ x_{\text{total}} = x_1 + x_2 + x_3 + \ldots + x_n \]
- According to Hooke's Law, the extension of each spring can be expressed as:
\[ x_i = \frac{F}{k} \]
- Thus, the total extension becomes:
\[ x_{\text{total}} = \frac{F}{k} + \frac{F}{k} + \ldots + \frac{F}{k} = n \cdot \frac{F}{k} \]
- Rearranging gives:
\[ x_{\text{total}} = \frac{nF}{k} \]
- The equivalent spring constant \( k_{\text{eq}} \) can be defined as:
\[ F = k_{\text{eq}} \cdot x_{\text{total}} \]
- Substituting for \( x_{\text{total}} \):
\[ F = k_{\text{eq}} \cdot \frac{nF}{k} \]
- Dividing both sides by \( F \) (assuming \( F \neq 0 \)):
\[ 1 = \frac{k_{\text{eq}} \cdot n}{k} \]
- Rearranging gives:
\[ k_{\text{eq}} = \frac{k}{n} \]

### Final Expressions:
- For the parallel combination of n springs:
\[ k_{\text{eq (parallel)}} = n \cdot k \]
- For the series combination of n springs:
\[ k_{\text{eq (series)}} = \frac{k}{n} \]