Oscillation: Practice Problem & Solution
A particle executes linear simple harmonic motion with an amplitude of $ 3 \text{ cm} $. When the particle is at $ 2 \text{ cm} $ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is: (2017-Delhi)
Solution Explained:
To solve this problem, we apply the core principles of Oscillation. Understanding the underlying formula is key to arriving at the correct answer below:
Given $ |v| = |a| $, we have $ \omega \sqrt{A^2 - x^2} = \omega^2 x $. Substituting $ A = 3 $ and $ x = 2 $, we get $ \sqrt{3^2 - 2^2} = \omega (2) \implies \omega = \frac{\sqrt{5}}{2} $. The time period is $ T = \frac{2\pi}{\omega} = \frac{4\pi}{\sqrt{5}} $.
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