A particle executes SHM, the graph of velocity as a function of displacement is
(a) a circle
(b) a p

Q: Why do we plot velocity against displacement instead of time?

Plotting velocity against displacement gives us a 'phase space' diagram. It's a powerful tool in physics because it shows the complete state of the system at any instant without explicitly depending on time.

Q: Can the graph ever be a circle?

Yes! If the angular frequency $\omega = 1 \text{ rad/s}$, the semi-major and semi-minor axes become equal (both equal to $A$), and the ellipse becomes a perfect circle.

Q: What do the intercepts on the axes represent?

The x-intercepts ($\pm A$) represent the extreme positions where velocity is zero. The v-intercepts ($\pm \omega A$) represent the maximum velocities which occur at the mean position ($x=0$).

Q: Why is the velocity positive and negative for the same displacement?

During one complete oscillation, the particle passes through the same position twice—once moving in the positive direction and once moving in the negative direction. This creates the top and bottom halves of the ellipse.

Q: Does this apply to damped harmonic motion?

No. In damped harmonic motion, energy is lost, so the amplitude decreases over time. The phase space graph would look like an inward spiraling curve rather than a closed ellipse.

Question

A particle executes SHM, the graph of velocity as a function of displacement is

Options:

(A) a circle

(B) a parabola

(C) an ellipse

(D) a helix

Sigma Percentile · Accepted Answer

### The Phase Space of Simple Harmonic Motion

When we study Simple Harmonic Motion (SHM), we usually focus on how displacement, velocity, or acceleration changes with time. However, there is another incredibly powerful way to visualize the motion: plotting velocity directly against displacement. This is known as a **phase space trajectory**.

Let's dive into the mathematics to see what shape this trajectory takes.

### The Master Equation

We start with the fundamental kinematic equation for a particle executing SHM. The velocity $v$ at any displacement $x$ from the mean position is given by:

$v = \omega \sqrt{A^2 - x^2}$

Here, $\omega$ is the angular frequency and $A$ is the maximum displacement, or amplitude. This equation tells us that velocity is maximum at the mean position ($x = 0$) and zero at the extreme positions ($x = \pm A$).

### Unveiling the Geometry

To understand the geometric shape this equation represents, we need to eliminate the square root. We do this by squaring both sides:

$v^2 = \omega^2 (A^2 - x^2)$

Now, let's expand the right side and bring all the variable terms ($x$ and $v$) to the left side of the equation:

$v^2 = \omega^2 A^2 - \omega^2 x^2$

$v^2 + \omega^2 x^2 = \omega^2 A^2$

This is starting to look like a familiar conic section. To make it perfectly match a standard mathematical form, we divide the entire equation by the constant term on the right, which is $\omega^2 A^2$:

$\frac{v^2}{\omega^2 A^2} + \frac{\omega^2 x^2}{\omega^2 A^2} = 1$

Notice how the $\omega^2$ cancels out in the second term. We are left with a beautifully symmetric equation:

$\frac{x^2}{A^2} + \frac{v^2}{(\omega A)^2} = 1$

### The Final Conclusion

Look closely at this final equation. It perfectly matches the standard equation of an **ellipse**:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

In our physical system, the $x$-axis represents displacement and the $y$-axis represents velocity. The semi-major and semi-minor axes of this ellipse are $A$ and $\omega A$.

Therefore, the graph of velocity as a function of displacement for a particle executing SHM is an **ellipse**.

**A Fun Thought Experiment:** What if the angular frequency $\omega$ happens to be exactly $1 	ext{ rad/s}$? In that special case, the denominators become equal ($A^2$), and the equation simplifies to $x^2 + v^2 = A^2$. The ellipse transforms into a perfect circle! But in the general case, it remains an ellipse.

Animated Solution for Physics - Oscillations: A particle executes SHM, the graph of velocity as a function of displacement is

Select Answer:

Visualized Solution

Velocity-Displacement Relation

Squaring the Equation

Rearranging Terms

Standard Form of Conic Section

Identifying the Graph

Special Case

The Sigma Insight: Simple Harmonic Motion (SHM)

The Phase Space of Simple Harmonic Motion

The Master Equation

Unveiling the Geometry

The Final Conclusion

Similar Questions

When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is

The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure. The potential energy $U (x)$ versus time ( $t$ ) plot of the particle is correctly shown in figure :

A particle executes simple harmonic motion represented by displacement function as $x (t) = A sin (ω t + ϕ)$ If the position and velocity of the particle at $t = 0$ s are 2 cm and $2 ω$ cm s $^{- 1}$ respectively, then its amplitude is $x 2$ cm, where the value of $x$ is ............ .

A particle executes simple harmonic motion between $x = - A$ and $x = + A$ . The time taken for it to go from $O$ to $A /2$ is $T_{1}$ and to go from $A /2$ to $A$ is $T_{2}$ , then

The $x - t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t = 4/3 s$ is

A particle moves with simple harmonic motion in a straight line. In first $τ$ sec, after starting from rest, it travels a distance $a$ and in next $τ$ sec, it travels $2 a$ in same direction, then

The displacement of an object attached to a spring and executing simple harmonic motion is given by $x = 2 \times 1 0^{- 2} cos π t$ metre. The time at which the maximum speed first occurs is

The function $sin^{2} (ω t)$ represents

A particle performs simple harmonic motion with amplitude $A$ . Its speed is trebled at the instant that it is at a distance $\frac{2}{3} A$ from equilibrium position. The new amplitude of the motion is