A particle executes simple harmonic motion and it is located at x = a, b and c at time t$$_0$$, 2t$$_0$$ and 3t$$_0$$ respectively. The frequency of the oscillation is:

We recall the standard equation of a particle executing simple harmonic motion (SHM):

$$x = A \cos(\omega t + \phi)$$

where $$A$$ is the amplitude, $$\omega$$ is the angular frequency and $$\phi$$ is the initial phase.

At three different instants we are given the displacements:

$$\begin{aligned} x &= a \quad\text{at}\; t = t_0,\\ x &= b \quad\text{at}\; t = 2t_0,\\ x &= c \quad\text{at}\; t = 3t_0. \end{aligned}$$

Substituting these data one by one in the SHM equation, we have

$$\begin{aligned} a &= A \cos\!\left(\omega t_0 + \phi\right),\\[4pt] b &= A \cos\!\left(2\omega t_0 + \phi\right),\\[4pt] c &= A \cos\!\left(3\omega t_0 + \phi\right). \end{aligned}$$

For convenience we introduce a new symbol

$$\theta = \omega t_0,$$

so that the three equations become

$$\begin{aligned} a &= A \cos\!\left(\phi + \theta\right),\\[4pt] b &= A \cos\!\left(\phi + 2\theta\right),\\[4pt] c &= A \cos\!\left(\phi + 3\theta\right). \end{aligned}$$

We now eliminate the unknowns $$A$$ and $$\phi$$. Dividing every equation by $$A$$ gives

$$\begin{aligned} \frac{a}{A} &= \cos(\phi + \theta),\\ \frac{b}{A} &= \cos(\phi + 2\theta),\\ \frac{c}{A} &= \cos(\phi + 3\theta). \end{aligned}$$

Next we use the trigonometric identity

$$\cos(\alpha + 2\beta) + \cos\alpha = 2\cos(\alpha + \beta)\,\cos\beta.$$

Setting $$\alpha = \phi + \theta$$ and $$\beta = \theta$$ in this identity, we obtain

$$\cos(\phi + 3\theta) + \cos(\phi + \theta) = 2\cos(\phi + 2\theta)\,\cos\theta.$$

Replacing the cosines by the corresponding ratios found above, this becomes

$$\frac{c}{A} + \frac{a}{A} = 2\left(\frac{b}{A}\right)\cos\theta.$$

Multiplying through by $$A$$ simplifies the expression to

$$a + c = 2b\,\cos\theta.$$

Solving for $$\cos\theta$$ we get

$$\cos\theta = \frac{a + c}{2b}.$$

Because $$\theta = \omega t_0,$$ we have

$$\theta = \cos^{-1}\!\left(\frac{a + c}{2b}\right).$$

Finally, the frequency $$f$$ is related to the angular frequency by $$\omega = 2\pi f$$, so

$$f = \frac{\omega}{2\pi} = \frac{\theta}{2\pi t_0} = \frac{1}{2\pi t_0}\,\cos^{-1}\!\left(\frac{a + c}{2b}\right).$$

Hence, the correct answer is Option A.