Question

A particle is executing S.H.M. between \(\mathrm{x}=-\mathrm{A}\) and \(\mathrm{x}=+\mathrm{A}\). If the time taken by the particle to travel from \(\mathrm{x}=0\) to \(\mathrm{A} / 2\) is \(\mathrm{T}_{1}\) and that taken to travel from \(\mathrm{x}=\mathrm{A} / 2\) to \(\mathrm{x}=\mathrm{A}\) is \(\mathrm{T}_{2}=\) then \(\ldots .\) (A) \(\mathrm{T}_{1}<\mathrm{T}_{2}\) (B) \(\mathrm{T}_{1}>\mathrm{T}_{2}\) (C) \(\mathrm{T}_{1}=2 \mathrm{~T}_{2}\) (D) \(\mathrm{T}_{1}=\mathrm{T}_{2}\)

Short answer

(A) \(T_{1}<T_{2}\)

Step-by-step solution

Review the Equation for SHM Position

In Simple Harmonic Motion, the position of a particle is given by the equation \(x(t) = A \sin{(\omega t)}\), where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(t\) is time. By knowing that the time taken to travel from an initial position to a final position can be found by comparing the corresponding sin values, we can proceed to the next step.

Calculate the Phase Angles for the Different Position

We need to find the phase angles that correspond to the positions \(x = 0, A/2\), and \(A\). For this, we can use the position equation of SHM. 1. For \(x = 0\), \(0 = A \sin{(\omega t_0)}\), \(\omega t_0 = 0\). 2. For \(x = A/2\), \(A/2 = A \sin{(\omega t_1)}\), \(\omega t_1 = \frac{\pi}{6}\) or \(\frac{5\pi}{6}\). 3. For \(x = A\), \(A = A \sin{(\omega t_2)}\), \(\omega t_2 = \frac{\pi}{2}\). For this exercise, we choose the smallest positive values \(\omega t_1'\) and \(\omega t_2'\) such that \(t_1\) < \(\omega t_1'\) < \(t_2\) < \( \omega t_2'\).

Compare T1 and T2

From the previous step, we can find the time \(T_1 = \frac{\pi}{6\omega} - \frac{0}{\omega} = \frac{\pi}{6\omega}\) and the time \(T_2 = \frac{\pi}{2\omega} - \frac{\pi}{6\omega} = \frac{\pi}{3\omega}\). Now, we can compare these times: Since \(\frac{\pi}{6\omega} < \frac{\pi}{3\omega}\), it means that \(T_1 < T_2\). Therefore, the correct answer is (A) \(\ T_{1}<\ T_{2}\).