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(C) The displacement of a particle in simple harmonic motion starting from an extreme position is given by $x(t) = A \cos(\omega t)$,where $A$ is the

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$\frac{1}{2\pi} \cos^{-1}\left[\frac{a+c}{2b}\right]$

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(C) The displacement of a particle in simple harmonic motion starting from an extreme position is given by $x(t) = A \cos(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.Let the displacements at times $t=1, 2, 3$ seconds be $a, b, c$ respectively.$a = A \cos(\omega)$$b = A \cos(2\omega)$$c = A \cos(3\omega)$Using the trigonometric identity $\cos(3	heta) + \cos(	heta) = 2 \cos(2	heta) \cos(	heta)$:$a + c = A \cos(\omega) + A \cos(3\omega) = A [\cos(\omega) + \cos(3\omega)]$$a + c = A [2 \cos(2\omega) \cos(\omega)]$Since $b = A \cos(2\omega)$,we can write:$a + c = 2b \cos(\omega)$$\cos(\omega) = \frac{a+c}{2b}$$\omega = \cos^{-1}\left[\frac{a+c}{2b}ight]$Since $\omega = 2\pi f$,where $f$ is the frequency:$f = \frac{1}{2\pi} \cos^{-1}\left[\frac{a+c}{2b}ight]$.

$A$ particle starts executing simple harmonic motion from one extreme position. If $a, b$ and $c$ are the displacements of the particle from the mean position at the ends of three successive seconds,the frequency of simple harmonic motion is

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