A body moves in SHM between two point 20m on the straight line Joining the points. If the angular speed of the…

1. Analyze the Question

We are given:

• The total path length (distance between the two extreme points) = 20 m.
• Angular speed ($\omega$) = 5 rad/s.
• Displacement from the center ($x$) = 6 m.
• We need to find the velocity ($v$) of the body at that specific displacement.

2. Explain the Concept

In Simple Harmonic Motion (SHM), the velocity $v$ of a body at any displacement $x$ from the equilibrium position (center) is given by the formula:

Where:

• $\omega$ is the angular speed.
• $A$ is the amplitude (the maximum displacement from the center).
• $x$ is the instantaneous displacement from the center.

Note: The amplitude ($A$) is half of the total path length because the body oscillates between $+A$ and $-A$.

3. Show the Solution

Step 1: Determine the amplitude ($A$). The body moves between two points 20 m apart. The distance from the center to either extreme point is the amplitude.

Step 2: Substitute the values into the velocity formula.

• $\omega = 5\text{ rad/s}$
• $A = 10\text{ m}$
• $x = 6\text{ m}$

4. Identify the Answer

Comparing our result to the given options: A. 20 m/s B. 30 m/s C. 40 m/s D. 50 m/s

The calculated value matches option C.

5. Exam Tip

Always remember that in SHM problems, the "distance between the two points" is the path length, which is $2A$. Students often mistake this for the amplitude ($A$) itself. Always divide by 2 before plugging it into your formula!