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. a/T
• B. 2a/T
• C. a/2T
• D. 3a/T

Hint:

Remember the time intervals: Extreme to a/2 takes T/6.

Answer 1

The Correct Option is D

Step 1: Set up the motion.
Measure the displacement of the SHM from the centre. Starting from an extreme point, $x = a\cos(\omega t)$, where $\omega = \dfrac{2\pi}{T}$. At $t = 0$ the particle is at the extreme, $x = a$.
Step 2: Distance the particle covers.
It moves from the extreme ($x = a$) to the point $x = \dfrac{a}{2}$, so the distance is $a - \dfrac{a}{2} = \dfrac{a}{2}$.
Step 3: Time for this travel.
Put $x = \dfrac{a}{2}$ in the equation: \[ \frac{a}{2} = a\cos(\omega t) \Rightarrow \cos(\omega t) = \frac{1}{2} \Rightarrow \omega t = \frac{\pi}{3}. \]
Step 4: Convert to time.
$t = \dfrac{\pi/3}{\omega} = \dfrac{\pi/3}{2\pi/T} = \dfrac{T}{6}$.
Step 5: Average velocity.
Average velocity is distance over time: \[ \bar v = \frac{a/2}{T/6} = \frac{a}{2}\times\frac{6}{T} = \frac{3a}{T}. \]
Step 6: Conclusion.
The magnitude of the average velocity is $\dfrac{3a}{T}$. \[ \boxed{\dfrac{3a}{T}} \]