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(D) Let the equation of motion be $x = A \cos(\omega t + \phi)$. However,for simplicity,we can use $x = A \cos(\omega t)$ if the phase is zero at $t=0

Vedclass · Accepted Answer

$\frac{1}{2\pi t_0} \cos^{-1} \left( \frac{a+c}{2b} \right)$

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(D) Let the equation of motion be $x = A \cos(\omega t + \phi)$. However,for simplicity,we can use $x = A \cos(\omega t)$ if the phase is zero at $t=0$. Given the positions at specific times:$a = A \cos(\omega t_0)$$b = A \cos(2\omega t_0)$$c = A \cos(3\omega t_0)$Using the trigonometric identity $\cos(3	heta) + \cos(	heta) = 2 \cos(2	heta) \cos(	heta)$:$a + c = A \cos(3\omega t_0) + A \cos(\omega t_0) = A [2 \cos(2\omega t_0) \cos(\omega t_0)]$Substitute $b = A \cos(2\omega t_0)$:$a + c = 2b \cos(\omega t_0)$$\cos(\omega t_0) = \frac{a+c}{2b}$$\omega t_0 = \cos^{-1} \left( \frac{a+c}{2b} ight)$Since $\omega = 2\pi f$,we have $2\pi f t_0 = \cos^{-1} \left( \frac{a+c}{2b} ight)$$f = \frac{1}{2\pi t_0} \cos^{-1} \left( \frac{a+c}{2b} ight)$

$A$ 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

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