Question

A particle starting from the origin executes simple harmonic motion along x-axis. Its velocity at any instant t is given by $v_{x}=22 \cos(\frac{\pi t}{2}) \text{ cm s}^{-1}$. The total distance covered by the particle in time $t=4.5 \text{ sec}$ is

• A. 74.5 cm
• B. 51.4 cm
• C. 65.9 cm
• D. 49.8 cm

Hint:

Total distance in one complete cycle $T$ is always equal to $4 \times \text{Amplitude}$.

Answer 1

The Correct Option is C

Step 1: Read the velocity equation.
We have $v_{x} = 22\cos\!\left(\dfrac{\pi t}{2}\right)$ cm s$^{-1}$. Compare with $v = v_{0}\cos(\omega t)$.

Step 2: Find $\omega$ and the time period.
So $\omega = \dfrac{\pi}{2}$ rad s$^{-1}$, and $T = \dfrac{2\pi}{\omega} = 4$ s. The biggest speed is $v_{0}=22$.

Step 3: Find the amplitude.
\[ A = \frac{v_{0}}{\omega} = \frac{22}{\pi/2} = \frac{44}{\pi} \approx 14.0 \text{ cm} \]
Step 4: Split the total time.
Total time is $4.5$ s. One full period is $4$ s, leaving $0.5$ s extra. In one full period the body covers $4A$.

Step 5: Distance in the extra $0.5$ s.
Starting from the centre, position is $x = A\sin(\omega t)$. At $t=0.5$ s, $x = A\sin\!\left(\dfrac{\pi}{4}\right) = \dfrac{A}{\sqrt{2}}$. That is the extra distance from the centre.

Step 6: Add it all up.
\[ \text{Distance} = 4A + \frac{A}{\sqrt{2}} = 14.0\,(4 + 0.707) \approx 65.9 \text{ cm} \]So the total distance is $65.9$ cm, which is option 3.
\[ \boxed{65.9 \text{ cm}} \]