If a simple harmonic oscillator has got a displacement of 0.02 m and acceleration equal to 2.0 m/s² at any time, the angular frequency of the oscillator is equal to

A simple harmonic oscillator has a displacement of \(0.02 \, \text{m}\) and acceleration equal to \(2.0 \, \text{m/s}^2\) at any time. What is the angular frequency of the oscillator?

(a) 10 rad/s

(b) 20 rad/s

(c) 30 rad/s

(d) 40 rad/s

Detailed Answer:

In Simple Harmonic Motion (SHM), the relationship between displacement \(x\), acceleration \(a\), and angular frequency \(\omega\) is given by the equation: \[ a = -\omega^2 x. \] Here: - \(x = 0.02 \, \text{m}\) is the displacement, - \(a = 2.0 \, \text{m/s}^2\) is the acceleration, - \(\omega\) is the angular frequency we need to calculate. Rearranging the equation to solve for \(\omega\): \[ \omega^2 = \frac{-a}{x}. \] Substituting the given values: \[ \omega^2 = \frac{-2.0}{0.02} = 100. \] Taking the square root of both sides: \[ \omega = \sqrt{100} = 10 \, \text{rad/s}. \] Therefore, the angular frequency of the oscillator is: \[ \boxed{10 \, \text{rad/s}} \ (\text{Option (a)}). \]

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