A particle executes simple harmonic motion and is located at ` x = a`, b at times `t_(0),2t_(0) and3t_(0)` respectively. The frequency of the oscillation is :

Options

A. `(1)/(2pit_(0))cos^(-1)((2a+3c)/(b))`
B. `(1)/(2pit_(0))cos^(-1)((a+c)/(2b))`
C. `(1)/(2pit_(0))cos^(-1)((a+b)/(2c))`
D. `(1)/(2pit_(0))cos^(-1)((a+2b)/(3c))`

Answer

To find the frequency of a particle executing simple harmonic motion (SHM) at given positions and times, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a particle in SHM located at positions \( A \), \( B \), and \( C \) at times \( t_0 \), \( 2t_0 \), and \( 3t_0 \) respectively. We need to find the frequency of the oscillation.

2. **Setting Up the Equations**:
The position of a particle in SHM can be described by the equation:
\[ x(t) = A \cos(\omega t + \phi) \]
For our specific times:
- At \( t = t_0 \): \( x(t_0) = A \cos(\omega t_0) = A \)
- At \( t = 2t_0 \): \( x(2t_0) = A \cos(2\omega t_0) = B \)
- At \( t = 3t_0 \): \( x(3t_0) = A \cos(3\omega t_0) = C \)

3. **Expressing Positions**:
We can express the positions at these times as:
\[ A = A \cos(\omega t_0) \quad (1) \]
\[ B = A \cos(2\omega t_0) \quad (2) \]
\[ C = A \cos(3\omega t_0) \quad (3) \]

4. **Adding Equations**:
Now, add equations (1) and (3):
\[ A + C = A \cos(\omega t_0) + A \cos(3\omega t_0) \]
Using the cosine addition formula:
\[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
We can rewrite it as:
\[ A + C = 2A \cos\left(2\omega t_0\right) \cos\left(\omega t_0\right) \]

5. **Relating B to A and C**:
From equation (2), we know:
\[ B = A \cos(2\omega t_0) \]
Therefore, we can express \( A + C \) in terms of \( B \):
\[ A + C = 2B \cos(\omega t_0) \]

6. **Finding \( \cos(\omega t_0) \)**:
Rearranging gives:
\[ \cos(\omega t_0) = \frac{A + C}{2B} \]

7. **Finding \( \omega \)**:
Now, substituting back into our expression for \( \omega t_0 \):
\[ \omega t_0 = \cos^{-1}\left(\frac{A + C}{2B}\right) \]

8. **Finding Frequency \( f \)**:
Recall that \( \omega = 2\pi f \), so:
\[ \omega = \frac{\cos^{-1}\left(\frac{A + C}{2B}\right)}{t_0} \]
Therefore, the frequency \( f \) can be expressed as:
\[ f = \frac{1}{2\pi} \cdot \frac{\cos^{-1}\left(\frac{A + C}{2B}\right)}{t_0} \]

### Final Result:
Thus, the frequency of the oscillation is:
\[ f = \frac{1}{2\pi t_0} \cos^{-1}\left(\frac{A + C}{2B}\right) \]