Question:
A particle executes simple harmonic motion with time period 'T' and amplitude 'a'. The magnitude of average velocity of the particle over the time interval during which it travels a distance from extreme position to a/2 is:
A particle executes simple harmonic motion with time period 'T' and amplitude 'a'. The magnitude of average velocity of the particle over the time interval during which it travels a distance from extreme position to a/2 is:
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Remember the time intervals: Extreme to a/2 takes T/6.
Updated On: Jun 6, 2026
- a/T
- 2a/T
- a/2T
- 3a/T
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The Correct Option is D
Solution and Explanation
Step 1: Position equation.
Starting from an extreme, $x = a\cos(\omega t)$ with $\omega = \dfrac{2\pi}{T}$.
Step 2: Distance covered.
From $x = a$ to $x = a/2$, the distance is $a/2$.
Step 3: Time taken.
$\dfrac{a}{2} = a\cos(\omega t) \Rightarrow \omega t = \dfrac{\pi}{3} \Rightarrow t = \dfrac{T}{6}$.
Step 4: Average velocity.
$\bar v = \dfrac{a/2}{T/6} = \dfrac{3a}{T}$. \[ \boxed{\dfrac{3a}{T}} \]
Starting from an extreme, $x = a\cos(\omega t)$ with $\omega = \dfrac{2\pi}{T}$.
Step 2: Distance covered.
From $x = a$ to $x = a/2$, the distance is $a/2$.
Step 3: Time taken.
$\dfrac{a}{2} = a\cos(\omega t) \Rightarrow \omega t = \dfrac{\pi}{3} \Rightarrow t = \dfrac{T}{6}$.
Step 4: Average velocity.
$\bar v = \dfrac{a/2}{T/6} = \dfrac{3a}{T}$. \[ \boxed{\dfrac{3a}{T}} \]
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