1. Write the SHM equation

For simple harmonic motion, $x(t)=A\cos(\omega t+\phi)$ where $A$ is amplitude, $\omega$ is angular frequency, and $\phi$ is phase constant.

2. Use the given positions

Given:

• at $t=t_0$, $x=a$
• at $t=2t_0$, $x=b$
• at $t=3t_0$, $x=c$

So, $A\cos(\omega t_0+\phi)=a$ $A\cos(2\omega t_0+\phi)=b$ $A\cos(3\omega t_0+\phi)=c$

Let $\theta=\omega t_0+\phi, \qquad \delta=\omega t_0$ Then the three equations become $A\cos\theta=a$ $A\cos(\theta+\delta)=b$ $A\cos(\theta+2\delta)=c$

3. Use the standard trigonometric identity

For cosine terms in arithmetic progression of angle, $\cos\theta+\cos(\theta+2\delta)=2\cos(\theta+\delta)\cos\delta$

Multiplying by $A$, $A\cos\theta + A\cos(\theta+2\delta)=2A\cos(\theta+\delta)\cos\delta$

Substitute $a,b,c$: $a+c=2b\cos\delta$

Hence, $\cos\delta=\frac{a+c}{2b}$

But $\delta=\omega t_0$, so $\omega t_0=\cos^{-1}\left(\frac{a+c}{2b}\right)$

Therefore, $\omega=\frac{1}{t_0}\cos^{-1}\left(\frac{a+c}{2b}\right)$

4. Convert angular frequency to frequency

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

5. Match with options

This matches Option A: $\boxed{\frac{1}{2\pi t_0}\cos^{-1}\left(\frac{a+c}{2b}\right)}$