The time period of a spring - mass system is T. If this spring is cut into two parts, whose lengths are in the ratio `1:2`, and the same mass is attached to the longer part, the new time period will be

Options

A. `sqrt((1)/(3))T`
B. `sqrt((2)/(3))T`
C. `(sqrt3T)/(2)`
D. `sqrt3T`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the initial time period of the spring-mass system
The time period \( T \) of a spring-mass system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass attached to the spring and \( k \) is the spring constant.

### Step 2: Cut the spring into two parts
The spring is cut into two parts in the ratio \( 1:2 \). Let the total length of the spring be \( L \). Then, the lengths of the two parts will be:
- Shorter part: \( \frac{L}{3} \)
- Longer part: \( \frac{2L}{3} \)

### Step 3: Determine the spring constant of the longer part
The spring constant \( k \) of a spring is inversely proportional to its length. If the original spring constant is \( k \) for length \( L \), then for the longer part of length \( \frac{2L}{3} \), the new spring constant \( k' \) can be calculated as follows:
\[ \frac{k'}{k} = \frac{L}{\frac{2L}{3}} \implies k' = k \cdot \frac{3}{2} = \frac{3k}{2} \]

### Step 4: Calculate the new time period
Now, we will find the new time period \( T' \) when the mass \( m \) is attached to the longer part of the spring:
\[ T' = 2\pi \sqrt{\frac{m}{k'}} \]
Substituting \( k' \):
\[ T' = 2\pi \sqrt{\frac{m}{\frac{3k}{2}}} = 2\pi \sqrt{\frac{2m}{3k}} = \sqrt{\frac{2}{3}} \cdot 2\pi \sqrt{\frac{m}{k}} = \sqrt{\frac{2}{3}} \cdot T \]

### Step 5: Conclusion
Thus, the new time period \( T' \) is:
\[ T' = \sqrt{\frac{2}{3}} T \]

### Final Answer
The new time period will be \( T' = \sqrt{\frac{2}{3}} T \).

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