A particle executes simple harmonic motion and is located at ` x = a`, b and c 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 oscillation of a particle executing simple harmonic motion (SHM) at specific positions at given times, we can follow these steps:

### Step-by-step Solution:

1. **Understanding the Motion**:
- The particle is located at positions \( x = a \), \( x = b \), and \( x = c \) at times \( t = t_0 \), \( t = 2t_0 \), and \( t = 3t_0 \) respectively.

2. **Using the SHM Equation**:
- The displacement in SHM can be represented as:
\[ x(t) = A \sin(\omega t) \]
- Here, \( A \) is the amplitude and \( \omega \) is the angular frequency.

3. **Setting Up the Equations**:
- At \( t = t_0 \):
\[ a = A \sin(\omega t_0) \quad \text{(1)} \]
- At \( t = 2t_0 \):
\[ b = A \sin(2\omega t_0) \quad \text{(2)} \]
- At \( t = 3t_0 \):
\[ c = A \sin(3\omega t_0) \quad \text{(3)} \]

4. **Using the Sine Addition Formula**:
- We can use the sine addition formula:
\[ \sin(a + b) = \sin a \cos b + \cos a \sin b \]
- For \( \sin(2\omega t_0) \) and \( \sin(3\omega t_0) \):
\[ \sin(2\omega t_0) = 2 \sin(\omega t_0) \cos(\omega t_0) \]
\[ \sin(3\omega t_0) = 3 \sin(\omega t_0) - 4 \sin^3(\omega t_0) \]

5. **Combining the Equations**:
- From equations (1), (2), and (3), we can express \( b \) in terms of \( a \) and \( c \):
\[ a + c = A \sin(\omega t_0) + A \sin(3\omega t_0) \]
- This leads to:
\[ a + c = 2A \sin(2\omega t_0) \cos(\omega t_0) \]

6. **Finding the Frequency**:
- Rearranging gives us:
\[ \cos(\omega t_0) = \frac{a + c}{2b} \]
- Now, using the relationship between angular frequency and frequency:
\[ \omega = 2\pi f \]
- Thus, we can express \( f \):
\[ \omega t_0 = \cos^{-1}\left(\frac{a + c}{2b}\right) \]
- Therefore, the frequency \( f \) can be expressed as:
\[ f = \frac{1}{2\pi t_0} \cos^{-1}\left(\frac{a + c}{2b}\right) \]

### Final Answer:
The frequency of oscillation \( f \) is given by:
\[ f = \frac{1}{2\pi t_0} \cos^{-1}\left(\frac{a + c}{2b}\right) \]