Q.A spring with one end attached to a mass and the other to a rigid support is stretched and released. (Note: more than one of the given options may be correct.)

The motion of a mass attached to a spring is an example of Simple Harmonic Motion (SHM).

1. Acceleration at release: When the spring is stretched and released, the mass is at its maximum displacement ($x = \pm A$, where $A$ is the amplitude). The acceleration in SHM is given by $a = -\omega^2 x$. Therefore, the magnitude of acceleration, $|a| = \omega^2 |x|$, is maximum when $|x|$ is maximum. So, (A) is correct.
2. Acceleration at equilibrium: At the equilibrium position, the displacement $x = 0$. Consequently, the acceleration $a = -\omega^2 (0) = 0$, which is its minimum value, not maximum. So, (B) is incorrect. …

When a mass attached to a spring is stretched and released, it undergoes Simple Harmonic Motion (SHM). This motion is characterized by a restoring force that is directly proportional to the displacement from the equilibrium position and acts in the opposite direction. The equilibrium position is where the net force on the mass is zero (the spring is neither stretched nor compressed from its natural length).

Let's define the key quantities in SHM:

• Displacement ($x$): The distance of the mass from its equilibrium position. It is maximum at the extreme points of oscillation (amplitude, $A$) and zero at the equilibrium position.
• Velocity ($v$): The rate of change of displacement.
• Acceleration ($a$): The rate of change of velocity.

The fundamental relationships for SHM are:

Let's analyze the behavior of acceleration and speed at different points in the motion:

• At the equilibrium position ($x=0$):

  • Acceleration: $a = -\omega^2 (0) = 0$. The magnitude of acceleration is minimum (zero).
  • Speed: $v = \pm \omega \sqrt{A^2 - 0^2} = \pm \omega A$. The magnitude of speed is maximum ($\omega A$).

• At the extreme positions ($x=\pm A$):

  • Acceleration: $a = -\omega^2 (\pm A) = \mp \omega^2 A$. The magnitude of acceleration is maximum ($\omega^2 A$).
  • Speed: $v = \pm \omega \sqrt{A^2 - A^2} = 0$. The magnitude of speed is minimum (zero).

Now, we can evaluate each option:

1. Option (A): Magnitude of acceleration, when just released is maximum.

   • "Just released" implies the mass is at one of its extreme positions, where its displacement from equilibrium is maximum ($x = \pm A$).
   • As derived above, at $x = \pm A$, the magnitude of acceleration is $|a| = \omega^2 A$, which is its maximum possible value.
   • This statement is correct.

2. Option (B): Magnitude of acceleration, when at equilibrium position, is maximum.

   • At the equilibrium position, $x=0$. …

Concept: SHM Extremes via Hooke's Law and Energy Conservation — No Sinusoidal Formula Needed

Method: Direct Force + Energy Reasoning (skip $a=-\omega^2x$ and $v=\omega\sqrt{A^2-x^2}$ entirely)

The stored answer plugs into the standard SHM formulas $a=-\omega^2x$ and $v=\pm\omega\sqrt{A^2-x^2}$. This method instead reasons directly from Hooke's law and energy conservation — the two physical facts those formulas are built from — reaching the same four verdicts without ever writing $\omega$, a sine, or a cosine.

Setup

Let the spring have constant $k$, the mass $m$, and let it be stretched to $x=A$ (the amplitude) before release from rest.

Steps

1. Acceleration from Hooke's law, directly. The restoring force is $F=-kx$, so by Newton's second law $a = F/m = -\dfrac{k}{m}x$. Its magnitude is $|a| = \dfrac{k}{m}|x|$ — a simple direct proportionality to the distance from equilibrium, nothing more.

2. (A) At release, $|x|=A$ (maximum possible displacement) — so by Step 1, $|a| = \dfrac{k}{m}A$, the largest value $|a|$ can ever take (since $|x|\le A$ always). (A) is correct.

3. (B) At equilibrium, $x=0$ — so $|a| = \dfrac{k}{m}(0) = 0$, the smallest possible value, not the largest. (B) is incorrect.

4. Speed from energy conservation, not from a velocity formula. Total mechanical energy is conserved: $\dfrac{1}{2}kx^2 + \dfrac{1}{2}mv^2 = \dfrac{1}{2}kA^2$ (all energy is spring PE at release, where $v=0$). Solving for speed:

This says speed is largest wherever $x^2$ is smallest — i.e. at $x=0$.

5. (C) At equilibrium, $x=0$, so by Step 4, $v^2 = \dfrac{k}{m}A^2$ — its maximum possible value (since $x^2\ge0$ can only ever subtract from $A^2$). (C) is correct. …