Amplitude of a harmonic oscillator – Rankers Physics

Equation of SHM: Practice Problem & Solution

The displacement of a harmonic oscillator is given by \(x = \alpha \sin\omega t + \beta \cos\omega t\). The amplitude of the oscillation is
\(\alpha + \beta\)
\(\alpha^2 + \beta^2\)
\(\sqrt{\alpha^2 + \beta^2}\)
\(\sqrt{\alpha^2 + \beta^2 + 2\alpha\beta}\)

Solution Explained:

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 represents two perpendicular SHMs of the same frequency with a phase difference of \(\frac{\pi}{2}\). The resultant amplitude is \(A = \sqrt{\alpha^2 + \beta^2}\).

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): Vibration of polyatomic molecules is not simple harmonic motion. Reason (R): The vibrations are superposition of SHMs of different frequency.  
2. Two Simple Harmonic Motions of angular frequency $100 \text{ rad s}^{-1}$ and $1000 \text{ rad s}^{-1}$ have the same displacement amplitude. The ratio of their maximum accelerations is: (2008)
3. Time period of a second's pendulum is \(2\text{ s}\), the approximate length of its string is equal to (\(g = 10\text{ m/s}^2\))
4. Equation of SHM of a particle whose amplitude is 0.1 m and frequency is 25 Hz with an initial phase of \(\frac{\pi}{4}\) radians is
5. A particle is performing simple harmonic motion of amplitude A and time period 12 seconds. If at t = 0 particle is at mean position, then distance travelled by particle in first 5 seconds will be: