AIPMT PRELIMS 2004 Physics Simple Harmonic Motion MCQ Question

In this question, we are examining the relationship between speed $v$ and acceleration $a$ of a particle executing simple harmonic motion (SHM). Let's break down the concepts involved and analyze each option.

Simple Harmonic Motion Overview

In SHM, the displacement $x$ of a particle can be described by the equation:

$$x(t) = A \cos(\omega t + \phi)$$

where:

• $A$ is the amplitude,
• $\omega$ is the angular frequency,
• $\phi$ is the phase constant.

The velocity $v$ of the particle is the first derivative of displacement with respect to time $t$:

$$v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi)$$

The acceleration $a$ of the particle is the first derivative of velocity with respect to time or the second derivative of displacement:

a(t) = \frac{dv}{dt} = -A \omega^2 \cos(\omega t + \phi) = -\omega^2 x(t)$$ ### Analyzing the Options Now, let's analyze each of the provided options: #### Option A: Value of $ a $ is zero, whatever may be the value of $ v $ This statement is **false**. In SHM, the acceleration $ a $ is not always zero. It varies with displacement $ x $. Specifically, $ a = -\omega^2 x $. When the particle is at the equilibrium position ($ x = 0 $), the acceleration is zero, but at other positions, it is not. #### Option B: When $ v $ is zero, $ a $ is zero This statement is **false**. In SHM, the velocity $ v $ is zero at the maximum displacement (the amplitude, $ A $), where the acceleration $ a $ is at its maximum value (given by $ a = -\omega^2 A $). Thus, $ v = 0 $ does not imply $ a = 0 $. #### Option C: When $ v $ is maximum, $ a $ is zero This statement is **true**. The velocity $ v $ reaches its maximum value when the particle passes through the equilibrium position ($ x = 0 $). At this point, the acceleration $ a $ is also zero because $ a = -\omega^2 x $ and $ x = 0 $. Thus, $ a = 0 $ when $ v $ is maximum. #### Option D: When $ v $ is maximum, $ a $ is maximum This statement is **false**. As mentioned, when the velocity $ v $ is maximum (at the equilibrium position), the acceleration $ a $ is zero. The acceleration reaches its maximum value at the amplitude position, where $ v $ is zero. ### Summary The correct answer is Option C: When $ v $ is maximum, $ a $ is zero. This reflects the fundamental characteristics of simple harmonic motion, where the velocity and acceleration exhibit sinusoidal relationships, and their maxima and minima occur at specific points in the motion cycle. ### Relevant Formulas 1. Displacement: $$ x(t) = A \cos(\omega t + \phi) $$ 2. Velocity: $$ v(t) = -A \omega \sin(\omega t + \phi) $$ 3. Acceleration: $$ a(t) = -A \omega^2 \cos(\omega t + \phi) = -\omega^2 x(t) $$ Understanding these relationships is crucial for solving problems related to oscillations and waves in physics.