Question:
If the maximum velocity in S.H.M. is 'V' then the average velocity during motion from one extreme to other extreme position will be
If the maximum velocity in S.H.M. is 'V' then the average velocity during motion from one extreme to other extreme position will be
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Average velocity is displacement divided by time; extreme to extreme takes half a period.
Updated On: Nov 12, 2026
- \(\frac{V}{\pi }\)
- \(\frac{2V}{\pi }\)
- \(\frac{4V}{\pi }\)
- \(\frac{8V}{\pi }\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
With amplitude \(A\) and angular frequency \(\omega\), maximum velocity is \(V=A\omega\).
Step 2: Displacement and time
From one extreme to the other: displacement \(=2A\), time \(=\dfrac{T}{2}=\dfrac{\pi}{\omega}\).
Step 3: Calculation
\(v_{avg}=\dfrac{2A}{\pi/\omega}=\dfrac{2A\omega}{\pi}=\dfrac{2V}{\pi}\)
Final Answer:
The average velocity is 2V divided by pi. \[ \boxed{\text{(B) }\dfrac{2V}{\pi}} \]
With amplitude \(A\) and angular frequency \(\omega\), maximum velocity is \(V=A\omega\).
Step 2: Displacement and time
From one extreme to the other: displacement \(=2A\), time \(=\dfrac{T}{2}=\dfrac{\pi}{\omega}\).
Step 3: Calculation
\(v_{avg}=\dfrac{2A}{\pi/\omega}=\dfrac{2A\omega}{\pi}=\dfrac{2V}{\pi}\)
Final Answer:
The average velocity is 2V divided by pi. \[ \boxed{\text{(B) }\dfrac{2V}{\pi}} \]
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