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Q13.5 (f):
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at 4 cm away from B going towards A.

Solution :

Step 1: Coordinate System Setup

The total distance between points $A$ and $B$ is $10\text{ cm}$. In Simple Harmonic Motion (SHM), the mean position ($O$) lies exactly at the center of the path $AB$.

  • Total length $2A = 10\text{ cm} \implies A = 5\text{ cm}$ (Amplitude).
  • Let the mean position $O$ be at coordinate $x = 0$.
  • Point $A$ is at $x = -5\text{ cm}$.
  • Point $B$ is at $x = +5\text{ cm}$.
A (-5) O (0) B (+5)

Step 2: Locating the Particle

The particle is $4\text{ cm}$ away from $B$ towards $A$. Since $B$ is at $x = +5\text{ cm}$, a position $4\text{ cm}$ from $B$ towards $A$ is calculated as:

$x = 5\text{ cm} - 4\text{ cm} = +1\text{ cm}$.

[By defining $B$ as the positive limit of oscillation].

Step 3: Determining the Sign of Velocity ($v$)

The problem states the particle is moving from $B$ towards $A$. The direction $A \rightarrow B$ is defined as positive. Therefore, the direction $B \rightarrow A$ is negative.

Since the particle is traveling towards $A$, the velocity vector points in the negative direction.

Sign of Velocity ($v$): Negative ($-$)

Step 4: Determining the Sign of Acceleration ($a$) and Force ($F$)

In SHM, the acceleration $a$ and the restoring force $F$ are governed by the relation:

$F = ma = -m\omega^2x$ [Per Hooke's Law for SHM].

At the current position, the displacement $x = +1\text{ cm}$.

  • Since $x$ is positive ($+1$), and the proportionality constant ($-m\omega^2$) is negative, the resulting acceleration $a$ must be negative.
  • Since $F = ma$, and $a$ is negative while mass $m$ is positive, the force $F$ must also be negative.

Summary Table of Signs

Parameter Sign Reasoning
Velocity ($v$) Negative ($-$) Particle is moving in the direction opposite to the defined positive $A \rightarrow B$.
Acceleration ($a$) Negative ($-$) $a = -\omega^2x$; since $x > 0$, $a < 0$.
Force ($F$) Negative ($-$) $F = ma$; force is directed towards the mean position (the negative direction).

Final Solution: The velocity is negative ($-$), acceleration is negative ($-$), and force is negative ($-$).


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