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

Key concepts

Amplitude in SHM

In simple harmonic motion (SHM), the amplitude is a crucial concept that defines the extent of oscillation. The amplitude, denoted as \( A \), is the maximum displacement of the particle from its mean position. For example, if a particle oscillates between \( x = -A \) and \( x = +A \), its amplitude is \( A \). This parameter is essential because it helps determine the maximum potential energy of the system and influences how far a particle will travel in its motion. In SHM, the particle’s position is a sinusoidal function of time, which can be described mathematically. Generally, this is given by the equation:

• \( x(t) = A \sin{(\omega t + \phi)} \)

Here \( \omega \) represents the angular frequency and \( \phi \) is the phase angle. It’s crucial to remember that the amplitude is always positive, as it represents a magnitude. Understanding amplitude helps in visualizing the limits within which the particle oscillates.

Angular frequency

Angular frequency, symbolized by \( \omega \), is a parameter that encapsulates how fast a particle oscillates in simple harmonic motion (SHM). It can be thought of as the rate of rotation in the angle of the oscillation. Expressed in radians per second, the angular frequency affects the speed of oscillation and is related to the period \( T \) of the SHM by the equation:

• \( \omega = \frac{2\pi}{T} \)

Angular frequency is crucial when performing calculations related to time intervals in SHM exercises, as it directly influences the expression \( \omega t \) used to determine phase angles. This parameter arises naturally from the oscillatory context and is a common factor in equations involving sinusoidal functions. Understanding \( \omega \) helps to solve problems related to the comparison of time intervals for different paths in SHM, as it determines how quickly a particle moves from one point to another in its cycle.

Phase angle in SHM

The phase angle in simple harmonic motion (SHM) defines the specific point within the oscillation cycle at a particular instant in time. Represented by \( \phi \) in the equation of motion \( x(t) = A \sin{(\omega t + \phi)} \), the phase angle shifts the sinusoidal wave along the time axis. This shift is vital when determining positions and times during the oscillation. In many problems, it helps us identify specific milestones within the cycle. For instance, when calculating the time taken for a particle to move between two positions, understanding and calculating phase angles become essential.

• For example, a phase angle of zero corresponds to the particle being at the equilibrium position at \( t = 0 \).
• If a phase angle is \( \frac{\pi}{2} \), the particle reaches its maximum amplitude at that instant.

Therefore, phase angles help us contextualize and predict a particle’s position at any time \( t \). Deeper understanding of phase angles enables us to analyze the time relationships between different segments of motion, resulting in solving exercises pertaining to comparing time intervals efficiently.