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.

• A. I, II and III
• B. II and III
• C. I and II
• D. I and III

Hint:

Velocity-displacement is elliptical; velocity-acceleration is elliptical; velocity-time is sinusoidal.

Answer 1

The Correct Option is B

Step 1: List the three claims.
We must judge three statements about graphs of a particle in simple harmonic motion: I velocity vs displacement is a parabola, II velocity vs time is sinusoidal, III velocity vs acceleration is an ellipse.

Step 2: Test statement I.
In SHM, velocity and displacement obey $v = \omega \sqrt{A^{2} - x^{2}}$. Squaring gives $\dfrac{v^{2}}{\omega^{2}} + x^{2} = A^{2}$. That is the equation of an ellipse, not a parabola. So statement I is wrong.

Step 3: Test statement II.
The velocity in SHM is $v = A \omega \cos(\omega t)$. A cosine of time is a sine type wave, so the velocity vs time graph is sinusoidal. Statement II is correct.

Step 4: Test statement III.
Here velocity is $v = A\omega \cos(\omega t)$ and acceleration is $a = -A\omega^{2} \sin(\omega t)$.

Step 5: Combine velocity and acceleration.
Using $\sin^{2} + \cos^{2} = 1$, we get $\left(\dfrac{v}{A\omega}\right)^{2} + \left(\dfrac{a}{A\omega^{2}}\right)^{2} = 1$. This is an ellipse, so statement III is correct.

Step 6: Pick the right combination.
Only II and III are correct. \[ \boxed{\text{II and III}} \]