Q188 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Physics
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
- A$\dfrac{V}{\pi}$
- Correct answerB$\dfrac{2V}{\pi}$
- C$\dfrac{4V}{\pi}$
- D$\dfrac{8V}{\pi}$
Solution
Maximum velocity in S.H.M. is given by $V = A\omega$, where $A$ is the amplitude and $\omega$ is the angular frequency.
The displacement of the particle from one extreme position to the other extreme position is $2A$.
The time taken to travel from one extreme to the other is half of the time period, which is $t = \dfrac{T}{2} = \dfrac{\pi}{\omega}$.
Average velocity is the ratio of total displacement to the total time taken.
$v_{avg} = \dfrac{2A}{\dfrac{\pi}{\omega}} = \dfrac{2A\omega}{\pi}$
Substituting $V = A\omega$, we get:
$v_{avg} = \dfrac{2V}{\pi}$
Answer: $\dfrac{2V}{\pi}$
The displacement of the particle from one extreme position to the other extreme position is $2A$.
The time taken to travel from one extreme to the other is half of the time period, which is $t = \dfrac{T}{2} = \dfrac{\pi}{\omega}$.
Average velocity is the ratio of total displacement to the total time taken.
$v_{avg} = \dfrac{2A}{\dfrac{\pi}{\omega}} = \dfrac{2A\omega}{\pi}$
Substituting $V = A\omega$, we get:
$v_{avg} = \dfrac{2V}{\pi}$
Answer: $\dfrac{2V}{\pi}$
Hence, the correct answer is option: 'B'.