Question

A particle executing SHM while moving from one extremity is found at distance $x_{1}, x_{2}$ and $x_{3}$ from the centre at the end of three successive seconds. The time period of oscillation is where $\theta=\cos ^{-1}\left(\frac{x_{1}+x_{3}}{2 x_{2}}\right)$

• A. $2 \pi /\theta$
• B. $\pi/ \theta$
• C. $\theta$
• D. $\pi / 2 \theta$

Answer 1

Answer: (A)

Displacement-time equation of the particle will be $X=A \cos \omega t$
Given that $x_{1}=A \cos \omega$
and $x_{2}=A \cos 2 \omega$
Now, $\frac{x_{1}+x_{3}}{2 x_{2}} =\frac{A(\cos \theta+\cos 3 \omega)}{2 A \cos 2 \omega} $
$=\frac{2 A \cos 2 \omega \cos \omega}{2 A \cos 2 \omega} $
$=\cos \omega $
$\omega =\cos ^{-1}\left(\frac{x_{1}+x_{2}}{2 x_{2}}\right)=\frac{2 \pi}{T}$
or $T=\frac{2 \pi}{\theta}$ where $\theta=\cos ^{-1}\left(\frac{x_{1}+x_{3}}{2 x_{2}}\right)$