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. \(\frac{1}{2\pi t_0}\cos^{-1}\left(\frac{a+c}{2b}\right)\)
• B. \(\frac{1}{2\pi t_0}\cos^{-1}\left(\frac{a+b}{2c}\right)\)
• C. \(\frac{1}{2\pi t_0}\cos^{-1}\left(\frac{2a+3c}{b}\right)\)
• D. \(\frac{1}{2\pi t_0}\cos^{-1}\left(\frac{a+2b}{3c}\right)\)

Hint:

Write \(x = A\cos(\omega t + \phi)\) and use \(\cos P + \cos Q = 2\cos\frac{P+Q}{2}\cos\frac{P-Q}{2}\) on the positions at \(t_0\) and \(3t_0\).

Answer 1

The Correct Option is A

Let \(x = A\cos(\omega t + \phi)\). Then \(a = A\cos(\omega t_0 + \phi)\), \(b = A\cos(2\omega t_0 + \phi)\), \(c = A\cos(3\omega t_0 + \phi)\).
Add the first and third: \(a + c = A\left[\cos(\omega t_0 + \phi) + \cos(3\omega t_0 + \phi)\right] = 2A\cos(2\omega t_0 + \phi)\cos(\omega t_0)\).
Since \(A\cos(2\omega t_0 + \phi) = b\): \(a + c = 2b\cos(\omega t_0)\).
So \(\omega t_0 = \cos^{-1}\left(\frac{a+c}{2b}\right)\) and \(\omega = \frac{1}{t_0}\cos^{-1}\left(\frac{a+c}{2b}\right)\).
Frequency \(f = \frac{\omega}{2\pi} = \frac{1}{2\pi t_0}\cos^{-1}\left(\frac{a+c}{2b}\right)\).
Answer: (1) \(\frac{1}{2\pi t_0}\cos^{-1}\left(\frac{a+c}{2b}\right)\)