A particle executes simple harmonic motion and is located at  x=a,b and  c at times to,3to and 5to respectively with respect to mean position . At  t=0, the particle is at the mean position and moving toward positive x-axis. The frequency of the oscillation is

# A particle executes simple harmonic motion and is located at  $x=a,b$ and  $c$ at times ${t}_{\text{o}},3{t}_{\text{o}}$ and $5{t}_{\text{o}}$ respectively with respect to mean position . At  $t=0$, the particle is at the mean position and moving toward positive x-axis. The frequency of the oscillation is

1. A

$\frac{1}{2\text{π}{t}_{\text{o}}}{\mathrm{cos}}^{-1}\left(\frac{a+b}{2c}\right)$

2. B

$\frac{1}{2\text{π}{t}_{\text{o}}}{\mathrm{cos}}^{-1}\left(\frac{a+b}{3c}\right)$

3. C

$\frac{1}{4\text{π}{t}_{\text{o}}}{\mathrm{cos}}^{-1}\left(\frac{a+c}{2b}\right)$

4. D

$\frac{1}{2\text{π}{t}_{\text{o}}}{\mathrm{cos}}^{-1}\left(\frac{a+c}{2b}\right)$

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### Solution:

$\begin{array}{l}⇒\omega =\frac{1}{2{t}_{0}}{\mathrm{cos}}^{-1}\left(\frac{a+c}{2b}\right)\left(\because \omega =2\pi f\right)\\ ⇒f=\frac{1}{4\pi {t}_{0}}×{\mathrm{cos}}^{-1}\left(\frac{a+c}{2b}\right)\end{array}$

Therefore, the correct answer is (C).

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