A particle is executing S.H.M. If`u_(1)` and `u_(2)` are the velocitiesof the particle at distances`x_(1)` and`x_(2)` from the mean position respectively, then

Options

A. `T=2pi sqrt((x_(1)+x_(2))/(u_(1)+u_(2)))`
B. `T=2pi sqrt((x_(2)^(2)+x_(1)^(2))/(u_(1)^(2)+u_(2)^(2)))`
C. `omega= sqrt((u_(2)^(2) - u_(1)^(2))/(x_(1)^(2) - x_(2)^(2))`
D. `2pi sqrt((u_(1)x_(2))/(u_(2)x_(1)))`

Answer

To solve the problem step by step, we will derive the relationship between the velocities of a particle executing Simple Harmonic Motion (SHM) at two different positions.

### Step 1: Understanding the velocities in SHM
In SHM, the velocity \( u \) of a particle at a distance \( x \) from the mean position can be expressed using the formula:
\[ u = \sqrt{a^2 - x^2} \]
where \( a \) is the amplitude of the motion.

### Step 2: Write the expressions for velocities at \( x_1 \) and \( x_2 \)
For the two positions \( x_1 \) and \( x_2 \), the velocities \( u_1 \) and \( u_2 \) can be expressed as:
\[ u_1 = \sqrt{a^2 - x_1^2} \]
\[ u_2 = \sqrt{a^2 - x_2^2} \]

### Step 3: Square both velocity equations
To eliminate the square roots, we square both expressions:
\[ u_1^2 = a^2 - x_1^2 \quad (1) \]
\[ u_2^2 = a^2 - x_2^2 \quad (2) \]

### Step 4: Rearranging the equations
From equation (1), we can express \( a^2 \):
\[ a^2 = u_1^2 + x_1^2 \quad (3) \]
From equation (2), we can also express \( a^2 \):
\[ a^2 = u_2^2 + x_2^2 \quad (4) \]

### Step 5: Equating the two expressions for \( a^2 \)
Since both equations (3) and (4) equal \( a^2 \), we can set them equal to each other:
\[ u_1^2 + x_1^2 = u_2^2 + x_2^2 \]

### Step 6: Rearranging to find the relationship
Rearranging the equation gives:
\[ u_2^2 - u_1^2 = x_1^2 - x_2^2 \]

### Step 7: Expressing in terms of angular frequency \( \omega \)
In SHM, the relationship between the angular frequency \( \omega \) and the displacements can be expressed as:
\[ \omega^2 = \frac{u_2^2 - u_1^2}{x_1^2 - x_2^2} \]

### Step 8: Final expression for \( \omega \)
Taking the square root gives us:
\[ \omega = \sqrt{\frac{u_2^2 - u_1^2}{x_1^2 - x_2^2}} \]

### Conclusion
Thus, the correct relationship is:
\[ \omega = \sqrt{\frac{u_2^2 - u_1^2}{x_1^2 - x_2^2}} \]