Which one of the following statements is correct regarding the provided displacement versus time curve for a particle executing simple harmonic motion?
Phase of the oscillating particle is same at t=1s and t=3s
Phase of the oscillating particle is same at t=2s and t=8s
Phase of the oscillating particle is same at t=3s and t=17s
Phase of the oscillating particle is same at t=4s and t=10s
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC NDA-1 – 2017
For a particle executing simple harmonic motion (SHM), the phase of oscillation at time $t$ is given by $\phi(t) = \omega t + \phi_0$, where $\omega = 2\pi/T$ is the angular frequency and $T$ is the period. The phase is the same at two different times $t_1$ and $t_2$ if their phase difference is an integer multiple of $2\pi$, i.e., $\phi(t_2) – \phi(t_1) = 2n\pi$, which simplifies to $\omega(t_2 – t_1) = 2n\pi$, or $(t_2 – t_1) = nT$, where $n$ is an integer. This means the phase is the same at times separated by an integer multiple of the period T. In option B, the time difference is $8s – 2s = 6s$. If the period T of the motion is a divisor of 6s (e.g., T=6s, T=3s, T=2s, T=1s), then 6s is an integer multiple of T, and the phase would be the same at these times. Assuming the question is valid and option B is the intended correct answer, it implies that 6s is an integer multiple of the period T, while the time differences in options A (2s) and C (14s) are not integer multiples of T. (Note: Option D also has a time difference of 6s, which could imply it is also correct if the period divides 6s. Without the actual graph, definitively determining the period and confirming only B is correct is not possible. However, based on the provided answer choices and typical SHM properties, a period that makes 6s a multiple is implied).
The phase of simple harmonic motion repeats itself every period T. Therefore, points in time separated by an integer multiple of T have the same phase.