Rankers Physics

Equation of SHM: Practice Problem & Solution

Identify the function which represents a periodic motion. (2020-Covid)
$ \log_e(\omega t) $
$ \sin \omega t + \cos \omega t $
$ e^{-\omega t} $
$ e^{\omega t} $

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 function $ \sin \omega t + \cos \omega t $ represents a superposition of two simple harmonic motions, making it periodic. The other given functions do not repeat their values over equal intervals of time.

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): Displacement-time equation of a particle moving along \(x\)-axis is \(x = 4 + 6 sin\omega t\). Under this situation, motion of particle is not simple harmonic. Reason (R): \(\frac{d^2x}{dt^2}\) for the given equation is not proportional to \(-x\).  
2. Assertion (A): In damped oscillation, the motion is periodic. Reason (R): In damped oscillation, the amplitude decreases due to dissipative forces.  
3. 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)
4. Assertion (A): Two particles are in SHM with same time period, same amplitude, same position and same speed are in the same phase. Reason (R): Phase of particle depends on position and speed of particle.  
5. A particle is performing SHM along x-axis such that its velocity and displacement are related as \(27v^2 = 10 - 3x^2\), then time period of oscillation of particle is: