Equation of SHM: Practice Problem & Solution
Assertion (A): Amplitude of SHM \(x = 4sin^2\omega t + 2cos^2\omega t + 2sin\omega t cos\omega t\) is \(sqrt{2}\). Reason (R): Angular frequency of given equation is \(2\omega\).
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Equation of SHM. Understanding the underlying formula is key to arriving at the correct answer below:
The expression \(x = 4sin^2\omega t + 2cos^2\omega t + 2sin\omega t cos\omega t\) simplifies to \(x = 3 + sin(2\omega t) - cos(2\omega t)\). The oscillatory part is \(sin(2\omega t) - cos(2\omega t)\). Assertion (A) is true, its amplitude is \(\sqrt{1^2 + (-1)^2} = \sqrt{2}\). Reason (R) is true, the angular frequency is \(2\omega\). However, the angular frequency does not explain the specific amplitude value.
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