A uniform spring whose unstressed length is l has a force constant k. The spring is cut into two pieces of unstressed lengths `l_(1)` and `l_(2)` where `l_(1)=nl_(2)` where n being on integer. Now a mass m is made to oscillate with first spring. The time period of its oscillation would be

Options

A. `T=2pisqrt((mn)/(k(n+1)))`
B. `T=2pisqrt((m)/(k(n+1)))`
C. `T=2pisqrt((m)/(k))`
D. `T=2pisqrt((m(n+1))/(nk))`

Explanation

To solve the problem step by step, we will analyze the situation involving the spring and the mass oscillating on it.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a uniform spring with an unstressed length \( l \) and a spring constant \( k \).
- The spring is cut into two pieces with lengths \( l_1 \) and \( l_2 \), where \( l_1 = n l_2 \) (with \( n \) being an integer).
- We need to find the time period of oscillation when a mass \( m \) is attached to the first spring \( l_1 \).

2. **Finding the Lengths of the Springs**:
- Since \( l_1 + l_2 = l \) and \( l_1 = n l_2 \), we can substitute to find \( l_2 \):
\[ n l_2 + l_2 = l \implies (n + 1) l_2 = l \implies l_2 = \frac{l}{n + 1} \]
- Now, substituting \( l_2 \) back to find \( l_1 \):
\[ l_1 = n l_2 = n \left(\frac{l}{n + 1}\right) = \frac{n l}{n + 1} \]

3. **Finding the Spring Constants**:
- The spring constant of a spring is inversely proportional to its length. The product of the length and spring constant remains constant for a uniform spring.
- For the original spring:
\[ k \cdot l = k_{\text{effective}} \cdot l_1 \implies k_{\text{effective}} = \frac{k \cdot l}{l_1} \]
- For the first spring \( l_1 \):
\[ k_{\text{effective}} = \frac{k \cdot l}{\frac{n l}{n + 1}} = \frac{(n + 1) k}{n} \]

4. **Calculating the Time Period**:
- The time period \( T \) of oscillation for a mass \( m \) attached to a spring is given by:
\[ T = 2 \pi \sqrt{\frac{m}{k_{\text{effective}}}} \]
- Substituting \( k_{\text{effective}} \):
\[ T = 2 \pi \sqrt{\frac{m}{\frac{(n + 1) k}{n}}} = 2 \pi \sqrt{\frac{m n}{(n + 1) k}} \]

5. **Final Result**:
- The time period of the oscillation of the mass \( m \) with the first spring \( l_1 \) is:
\[ T = 2 \pi \sqrt{\frac{m n}{(n + 1) k}} \]