The Question is from neet-physics motion-in-a-straight-line , and it was created by the author Mr. ANIL KUMAR.G

Question

A particle undergoing simple harmonic motion has an amplitude $A$ and angular frequency $\omega$. What is its average speed over one complete oscillation?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [Option C]
## CONCEPT FRAMEWORK
	**Primary Concept**: Average Speed in Simple Harmonic Motion
	**Related Topics**: Simple Harmonic Motion, Amplitude, Angular Frequency, Period
	**Prerequisites**: Understanding of simple harmonic motion, concept of average speed, relationship between angular frequency and period.
## SOLUTION PATHWAY
**Given Information**:
	Amplitude of SHM = $A$
	Angular frequency of SHM = $\omega$
**Method & Steps**:
	1. Calculate the total distance traveled in one complete oscillation.
		Reasoning: In one complete oscillation, the particle moves from the equilibrium position to one extreme, then to the other extreme, and back to the equilibrium position. This is a distance of 4 times the amplitude.
		Formula/Rule: Total distance = 4A
	2. Calculate the period of oscillation.
		Reasoning: The period is related to angular frequency by the formula: $T = \frac{2\pi}{\omega}$
		Formula/Rule: $T = \frac{2\pi}{\omega}$
	3. Calculate the average speed.
		Reasoning: Average speed is the total distance traveled divided by the time taken. In this case, the time is the period of oscillation.
		Formula/Rule: Average speed = Total distance / Time
		Working: Average speed = $\frac{4A}{2\pi/\omega} = \frac{4A\omega}{2\pi}$
		Key Point: Average speed is the total distance divided by total time, not the displacement divided by time.
	4.  Final Result
		Result: Average speed = $\frac{4A\omega}{2\pi}$
		Verification: The units match (distance/time). The result is consistent with SHM theory.
 ## OPTION ANALYSIS
[A] :x: [This is the maximum speed of the particle in SHM, not the average speed.]
[B] :x: [This is twice the maximum speed, and it's not the average speed.]
[C] ✓: [Correct, derived in the solution pathway]
[D] :x: [Average speed cannot be zero for a moving particle.]
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Using maximum speed instead of average speed]: [Always consider the definition of average speed as total distance over total time.]
		2.[Confusing displacement with distance]: [Displacement in one complete oscillation is zero, but the distance is not.]
	**Quick Tips**:
		1.[Remember the total distance in one SHM cycle is 4A]
		2.[Always use the period $T=\frac{2\pi}{\omega}$ for a complete oscillation when calculating average speed.]

Mr. ANIL KUMAR.G · Answer

$A\omega$

Mr. ANIL KUMAR.G · Answer

$2A\omega$

Mr. ANIL KUMAR.G · Answer

The correct Option is "$4A\omega/2\pi$"

Mr. ANIL KUMAR.G · Answer

Zero

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