A block of mass `M` suspended from a spring oscillates with time period `T`. If spring is cut in to two equal parts and same mass `M` is suspended from one part, new period os oscillation is

Options

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

Explanation

To solve the problem step by step, we will analyze the situation of the spring and the mass in detail.

### Step 1: Understand the initial conditions
We have a block of mass \( M \) suspended from a spring with a time period \( T \). The time period of a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{M}{k}} \]
where \( k \) is the spring constant.

### Step 2: Cut the spring into two equal parts
When the spring is cut into two equal parts, each part will have a new length of \( \frac{L}{2} \) (where \( L \) is the original length of the spring).

### Step 3: Determine the new spring constant
The spring constant of a spring is inversely proportional to its length. When the length of the spring is halved, the spring constant doubles. Therefore, if the original spring constant is \( k \), the new spring constant \( k' \) for each half is:
\[ k' = 2k \]

### Step 4: Calculate the new time period
Now, we will find the new time period \( T' \) when the mass \( M \) is suspended from one of the cut springs. The time period for the new spring system is given by:
\[ T' = 2\pi \sqrt{\frac{M}{k'}} \]
Substituting \( k' = 2k \) into the equation gives:
\[ T' = 2\pi \sqrt{\frac{M}{2k}} \]

### Step 5: Relate the new time period to the original time period
We can factor out the \( \sqrt{2} \) from the equation:
\[ T' = 2\pi \sqrt{\frac{M}{k}} \cdot \frac{1}{\sqrt{2}} = \frac{T}{\sqrt{2}} \]
where \( T = 2\pi \sqrt{\frac{M}{k}} \) is the original time period.

### Conclusion
Thus, the new time period of oscillation when the mass \( M \) is suspended from one part of the spring is:
\[ T' = \frac{T}{\sqrt{2}} \]