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
$a +c=A \cos \omega t_{0}+A \cos 3 \omega t_{0}$
$\Rightarrow a+ c=A\left[\cos \omega t_{0}+\cos 3 \omega t_{0}\right]$
$\Rightarrow (a +c)=2 A\left[\frac{\cos \left(3 \omega t_{0}-\omega t_{0}\right)}{2}\right] \cos \left(\frac{3 \omega t_{0}+\omega t_{0}}{2}\right)$
$\Rightarrow (a +c)=2 A \cos \omega t_{0} \cos 2 \omega t_{0}$
Using $\cos C+\cos D=2 \cos \frac{(C-D)}{2} \cos \frac{(C+D)}{2}$]
$\Rightarrow a +c=2 b \cos \omega t_{0}$
$\Rightarrow \cos \omega t_{0}=\frac{a+c}{2 b}$
$\Rightarrow \omega t_{0}=\cos ^{-1}\left(\frac{a+ c}{2 b}\right) \\ 2 \pi f t_{0}=\cos ^{-1}\left(\frac{a+c}{2 b}\right) \text { or } f=\frac{1}{2 \pi t_{0}} \cos ^{-1}\left(\frac{a +c}{2 b}\right)$