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:

Since the two perpendicular components have a phase difference of \(\frac{\pi}{2}\), the net amplitude is \(A = \sqrt{\alpha^2 + \beta^2}\).

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): In damped oscillation both amplitude and frequency change with time. Reason (R): Both amplitude and frequency vary exponentially.
2. A vertical spring block system is made to oscillate. Assertion (A): Its time period on earth is more than that on the moon. Reason (R): Its extension on moon (in equilibrium) is more than that on the earth.  
3. A particle is executing a simple harmonic motion. Its maximum acceleration is $ \alpha $ and maximum velocity is $ \beta $. Then, its time period of vibration will be: (2015 Re)
4. A body oscillates in SHM according to the equation \( x = 10 cos \left(2\pi t + \frac{\pi}{4} \right)\text{ cm}\). Its instantaneous displacement at \(t = 1\text{ s}\) is
5. The particle executing simple harmonic motion has a kinetic energy $K_0 \cos^2 \omega t$. The maximum values of the potential energy and the total energy are respectively: (2007)