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 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
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Total distance in one complete cycle $T$ is always equal to $4 \times \text{Amplitude}$.
Updated On: Jun 3, 2026
- 74.5 cm
- 51.4 cm
- 65.9 cm
- 49.8 cm
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The Correct Option is C
Solution and Explanation
Step 1: Concept
The velocity of a particle executing simple harmonic motion starting from the origin is expressed as $v(t) = v_0 \cos(\omega t)$, where $v_0 = \omega A$ is the maximum velocity and $A$ is the amplitude.
Step 2: Meaning
Comparing the given equation $v_{x}=22 \cos(\frac{\pi t}{2})$ with the standard form, we find the angular frequency $\omega = \frac{\pi}{2} \text{ rad s}^{-1}$ and maximum velocity $v_0 = 22 \text{ cm s}^{-1}$. Thus, the time period is $T = \frac{2\pi}{\omega} = 4 \text{ seconds}$, and the amplitude is $A = \frac{v_0}{\omega} = \frac{22}{\pi/2} = \frac{44}{\pi} \text{ cm}$.
Step 3: Analysis
In one complete time period ($T = 4 \text{ s}$), the total distance covered by a particle executing SHM is $4A$. The remaining time is $4.5 - 4 = 0.5 \text{ seconds}$, which is equivalent to $\frac{1}{8}$ of a period ($T/8$). Since the particle moves from the mean position to $x = A \sin(\omega t)$, at $t = 0.5 \text{ s}$, the additional distance covered is $A \sin(\frac{\pi}{2} \times 0.5) = A \sin(\frac{\pi}{4}) = \frac{A}{\sqrt{2}}$. The total distance is $4A + \frac{A}{\sqrt{2}} = A(4 + \frac{1}{\sqrt{2}}) = \frac{44}{\pi}(4 + 0.7071) \approx 14.0056 \times 4.7071 \approx 65.9 \text{ cm}$.
Step 4: Conclusion
Re-evaluating the correct key match from standard reference calculations ensures the absolute geometric distance yields 65.9 cm.
Final Answer: (C)
The velocity of a particle executing simple harmonic motion starting from the origin is expressed as $v(t) = v_0 \cos(\omega t)$, where $v_0 = \omega A$ is the maximum velocity and $A$ is the amplitude.
Step 2: Meaning
Comparing the given equation $v_{x}=22 \cos(\frac{\pi t}{2})$ with the standard form, we find the angular frequency $\omega = \frac{\pi}{2} \text{ rad s}^{-1}$ and maximum velocity $v_0 = 22 \text{ cm s}^{-1}$. Thus, the time period is $T = \frac{2\pi}{\omega} = 4 \text{ seconds}$, and the amplitude is $A = \frac{v_0}{\omega} = \frac{22}{\pi/2} = \frac{44}{\pi} \text{ cm}$.
Step 3: Analysis
In one complete time period ($T = 4 \text{ s}$), the total distance covered by a particle executing SHM is $4A$. The remaining time is $4.5 - 4 = 0.5 \text{ seconds}$, which is equivalent to $\frac{1}{8}$ of a period ($T/8$). Since the particle moves from the mean position to $x = A \sin(\omega t)$, at $t = 0.5 \text{ s}$, the additional distance covered is $A \sin(\frac{\pi}{2} \times 0.5) = A \sin(\frac{\pi}{4}) = \frac{A}{\sqrt{2}}$. The total distance is $4A + \frac{A}{\sqrt{2}} = A(4 + \frac{1}{\sqrt{2}}) = \frac{44}{\pi}(4 + 0.7071) \approx 14.0056 \times 4.7071 \approx 65.9 \text{ cm}$.
Step 4: Conclusion
Re-evaluating the correct key match from standard reference calculations ensures the absolute geometric distance yields 65.9 cm.
Final Answer: (C)
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