A particle is vibrating in SHM. If its velocities are `v_1` and `v_2` when the displacements from the mean postion are `y_1` and `y_2`, respectively, then its time period is

Options

A. `2pisqrt((y_1^2+y_2^2)/(v_1^2+v_2^2))`
B. `2pisqrt((v_1^2+v_2^2)/(y_1^2+y_2^2))`
C. `2pisqrt((v_1^2-v_2^2)/(y_1^2-y_2^2))`
D. `2pisqrt((y_1^2-y_2^2)/(v_1^2-v_2^2))`

Explanation

To find the time period of a particle vibrating in Simple Harmonic Motion (SHM) given its velocities and displacements, we can follow these steps:

### Step 1: Understand the relationship between velocity, displacement, and angular frequency
In SHM, the velocity \( v \) of a particle at a displacement \( y \) from the mean position is given by the formula:
\[ v = \omega \sqrt{A^2 - y^2} \]
where:
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude of the motion,
- \( y \) is the displacement from the mean position.

### Step 2: Write the equations for the two given conditions
We have two conditions given in the problem:
1. When the displacement is \( y_1 \), the velocity is \( v_1 \):
\[ v_1 = \omega \sqrt{A^2 - y_1^2} \]
2. When the displacement is \( y_2 \), the velocity is \( v_2 \):
\[ v_2 = \omega \sqrt{A^2 - y_2^2} \]

### Step 3: Square both equations to eliminate the square root
Squaring both equations gives us:
1. \( v_1^2 = \omega^2 (A^2 - y_1^2) \)
2. \( v_2^2 = \omega^2 (A^2 - y_2^2) \)

### Step 4: Rearrange both equations to express \( A^2 \)
From the first equation:
\[ A^2 = \frac{v_1^2}{\omega^2} + y_1^2 \]
From the second equation:
\[ A^2 = \frac{v_2^2}{\omega^2} + y_2^2 \]

### Step 5: Set the two expressions for \( A^2 \) equal to each other
Since both expressions are equal to \( A^2 \), we can set them equal:
\[ \frac{v_1^2}{\omega^2} + y_1^2 = \frac{v_2^2}{\omega^2} + y_2^2 \]

### Step 6: Rearrange to solve for \( \omega^2 \)
Rearranging gives us:
\[ \frac{v_1^2 - v_2^2}{\omega^2} = y_2^2 - y_1^2 \]
Thus,
\[ \omega^2 = \frac{v_1^2 - v_2^2}{y_2^2 - y_1^2} \]

### Step 7: Find the time period \( T \)
The time period \( T \) is related to angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting for \( \omega \) gives:
\[ T = 2\pi \sqrt{\frac{y_2^2 - y_1^2}{v_1^2 - v_2^2}} \]

### Final Answer
Thus, the time period \( T \) of the particle's motion is:
\[ T = 2\pi \sqrt{\frac{y_2^2 - y_1^2}{v_1^2 - v_2^2}} \]