Question

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 :

• A. $\frac{1}{2\pi t_o}{\cos }^{-1}\left(\frac{a+c}{2b}\right)$
• B. $\frac{1}{2\pi t_o}{\cos }^{-1}\left(\frac{a+b}{2c}\right)$
• C. $\frac{1}{2\pi t_o}{\cos }^{-1}\left(\frac{2a+3c}{b}\right)$
• D. $\frac{1}{2\pi t_o}{\cos }^{-1}\left(\frac{a+2b}{3c}\right)$

Answer 1

Answer: (A)

The simple harmonic motion can be represented by
$a=A\cos \omega t_0$
$b=A\cos 2\sigma \omega t_0$
$c=A\cos 3\omega t_0$
Adding Eqs. (1) and (3), we get
$a+c=A\cos \omega t_0+A\cos 3\omega t_0$
$\Rightarrow a+c=A\left[\cos \omega t_0+\cos 3\omega t_0\right]$
$\Rightarrow (a+c)=2A\left[\frac{\cos \left(3\omega t_0-\omega t_0\right)}{2}\right]\cos \left(\frac{3\omega t_0+\omega t_0}{2}\right)$
$\Rightarrow (a+c)=2A\cos \omega t_0\cos 2\omega t_0$
Using $\cos C+\cos D=2\cos \frac{(C-D)}{2}\cos \frac{(C+D)}{2}$]
$\Rightarrow a+c=2b\cos \omega t_0$
$\Rightarrow \cos \omega t_0=\frac{a+c}{2b}$
$$\begin{gathered} \Rightarrow \omega t_0={\cos }^{-1}\left(\frac{a+c}{2b}\right) \\ 2\pi f{t}_0={\cos }^{-1}\left(\frac{a+c}{2b}\right)\text{ or }f=\frac{1}{2\pi t_0}{\cos }^{-1}\left(\frac{a+c}{2b}\right) \end{gathered}$$