Question:
Which of the following statements are correct for a particle in simple harmonic motion?
I. Its velocity-displacement graph is a parabola.
II. Its velocity-time graph is sinusoidal.
III. Its velocity-acceleration graph is an ellipse.
Which of the following statements are correct for a particle in simple harmonic motion?
I. Its velocity-displacement graph is a parabola.
II. Its velocity-time graph is sinusoidal.
III. Its velocity-acceleration graph is an ellipse.
Show Hint
Velocity-displacement is elliptical; velocity-acceleration is elliptical; velocity-time is sinusoidal.
Updated On: Jun 10, 2026
- I, II and III
- II and III
- I and II
- I and III
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Concept
SHM kinematics and graphical representation.
Step 2: Analysis
* Statement I: Velocity $v = \omega \sqrt{A^2 - x^2}$. Squaring gives $v^2/\omega^2 + x^2 = A^2$, which is an ellipse, not a parabola. * Statement II: $v = A\omega \cos(\omega t)$, which is sinusoidal. * Statement III: $v = A\omega \cos(\omega t)$ and $a = -A\omega^2 \sin(\omega t)$. Since $\sin^2\theta + \cos^2\theta = 1$, $(v/A\omega)^2 + (a/A\omega^2)^2 = 1$, which is an ellipse.
Step 3: Conclusion
Statements II and III are correct.
Final Answer: (B)
SHM kinematics and graphical representation.
Step 2: Analysis
* Statement I: Velocity $v = \omega \sqrt{A^2 - x^2}$. Squaring gives $v^2/\omega^2 + x^2 = A^2$, which is an ellipse, not a parabola. * Statement II: $v = A\omega \cos(\omega t)$, which is sinusoidal. * Statement III: $v = A\omega \cos(\omega t)$ and $a = -A\omega^2 \sin(\omega t)$. Since $\sin^2\theta + \cos^2\theta = 1$, $(v/A\omega)^2 + (a/A\omega^2)^2 = 1$, which is an ellipse.
Step 3: Conclusion
Statements II and III are correct.
Final Answer: (B)
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