A particle executes simple harmonic motion and is located at ` x = a, b` and `c` at times `t_(0), 2t_(0)` and `3t_(0)` respectively. The frequency of the oscillation is :

Options

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

Answer

To solve the problem, we need to analyze the positions of the particle executing simple harmonic motion (SHM) at different times and relate them to the frequency of oscillation.

### Step-by-Step Solution:

1. **Understanding SHM**:
The position of a particle in simple harmonic motion can be described by the equation:
\[ x(t) = A \cos(\omega t + \phi) \]
where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant.

2. **Positions at Given Times**:
We know the positions at specific times:
- At \(t = t_0\), \(x = a\)
- At \(t = 2t_0\), \(x = b\)
- At \(t = 3t_0\), \(x = c\)

Thus, we can write:
\[ a = A \cos(\omega t_0 + \phi) \]
\[ b = A \cos(2\omega t_0 + \phi) \]
\[ c = A \cos(3\omega t_0 + \phi) \]

3. **Using Cosine Addition**:
We can derive a relationship using the cosine addition formula. Adding the equations for \(a\) and \(c\):
\[ a + c = A \cos(\omega t_0 + \phi) + A \cos(3\omega t_0 + \phi) \]
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 express this as:
\[ a + c = 2A \cos\left(2\omega t_0 + \phi\right) \cos\left(\omega t_0\right) \]

4. **Relating \(b\) to \(a\) and \(c\)**:
Given that \(b = 2a\), we can substitute this into our equations:
\[ 2a = A \cos(2\omega t_0 + \phi) \]

5. **Substituting \(b\) into the equation**:
Now we can express \(b\) in terms of \(a\) and \(c\):
\[ 2a = A \cos(2\omega t_0 + \phi) \]

6. **Finding \(\omega\)**:
From the above equations, we can derive:
\[ \cos(2\omega t_0 + \phi) = \frac{2a}{A} \]
Using the earlier relationship:
\[ a + c = 2A \cos(2\omega t_0 + \phi) \cos(\omega t_0) \]
We can express \(\cos(\omega t_0)\) in terms of \(a\) and \(c\):
\[ \cos(\omega t_0) = \frac{a + c}{2A} \]

7. **Finding Frequency**:
The angular frequency \(\omega\) is related to frequency \(f\) by:
\[ \omega = 2\pi f \]
Thus, we can express the frequency in terms of \(t_0\):
\[ f = \frac{1}{2\pi t_0} \cos^{-1}\left(\frac{a + c}{2b}\right) \]

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