A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x1(t) = A sin

Q: Why can we treat simple harmonic motion displacements as vectors?

Simple harmonic motion can be viewed as the projection of uniform circular motion onto a diameter. The rotating vector representing this circular motion is called a phasor. Since the frequency $\omega$ is the same for all waves, their relative phase angles remain constant, allowing us to add them using standard 2D vector addition.

Q: What if the frequencies of the two waves were different? Could we still use the phasor method?

No! The phasor method relies on the relative phase difference remaining constant over time. If the frequencies are different, the phasors rotate at different speeds, meaning the angle between them changes continuously, and they do not form a simple static vector diagram.

Q: Why does the resultant of two equal vectors of magnitude $A$ at $120^\circ$ have a magnitude of exactly $A$?

This is a beautiful geometric property! If you place two equal vectors of length $A$ at $120^\circ$ head-to-tail, they form two sides of an equilateral triangle. The third side, which represents the resultant, must also have a length of $A$. Mathematically, $\cos(120^\circ) = -0.5$, so $A^2 + A^2 + 2A^2(-0.5) = A^2$, giving a resultant of $A$.

Q: Why did we add $\pi$ to the phase angle of the resultant to find $\phi$?

To completely cancel out a vector, you need a vector of the same magnitude pointing in the exact opposite direction. In a polar coordinate system, pointing in the opposite direction corresponds to adding $180^\circ$ or $\pi$ radians to the angle.

Q: Is the algebraic method using trigonometric identities always more accurate than the phasor method?

Both methods are mathematically rigorous and yield the exact same result. However, the algebraic method involves tedious expansions of $\sin(\omega t + \theta)$ and solving simultaneous equations, which increases the chance of silly mistakes under exam pressure. The phasor method is highly visual, intuitive, and much faster.

Question

A point mass is subjected to two simultaneous sinusoidal displacements in $x$-direction, $x_1(t) = A \sin \omega t$ and $x_2(t) = A \sin \left(\omega t + \frac{2\pi}{3}\right)$. Adding a third sinusoidal displacement $x_3(t) = B \sin (\omega t + \phi)$ brings the mass to a complete rest. The values of $B$ and $\phi$ are

Options:

(A) $\sqrt{2}A, \frac{3\pi}{4}$

(B) $A, \frac{4\pi}{3}$

(C) $\sqrt{3}A, \frac{5\pi}{6}$

(D) $A, \frac{\pi}{3}$

Sigma Percentile · Accepted Answer

### Introduction to Superposition and Phasors

Imagine a calm pond. If you drop a pebble, ripples spread outward in beautiful concentric circles.

Now, imagine dropping two pebbles simultaneously at different spots. Where the ripples meet, they don't crash and shatter like solid objects; instead, they pass through each other, momentarily creating a new wave pattern that is the exact sum of the individual ripples.

This is the **Principle of Superposition**, one of the most fundamental and elegant concepts in wave mechanics and Simple Harmonic Motion (SHM).

In this problem, we are dealing with a point mass subjected to three simultaneous sinusoidal displacements along the same line.

Our goal is to find the parameters of the third displacement such that the net effect is absolute silence—the mass is brought to a complete rest.

$$
x_1(t) + x_2(t) + x_3(t) = 0
$$

While we can solve this using heavy trigonometric algebra, there is a much more elegant, visual, and intuitive tool at our disposal: **Phasors**.

---

### The Magic of Phasors

How do we represent a one-dimensional oscillation as a two-dimensional vector?

Recall that Simple Harmonic Motion is simply the projection of uniform circular motion onto a straight line.

If a vector of length $A$ rotates in a circle with a constant angular velocity $\omega$, its projection on the vertical axis is given by $A \sin \omega t$.

This rotating vector is called a **phasor**.

By representing sinusoidal functions as phasors, we can turn a tedious trigonometry problem into a simple vector addition problem!

Let's map our given displacements to phasors:

1. The first displacement is $x_1(t) = A \sin \omega t$. Its phasor $\vec{A}_1$ has a magnitude of $A$ and lies along the positive $x$-axis (phase angle $0$).

2. The second displacement is $x_2(t) = A \sin \left(\omega t + \frac{2\pi}{3}\right)$. Its phasor $\vec{A}_2$ has a magnitude of $A$ and is at an angle of $\frac{2\pi}{3}$ ($120^\circ$) relative to the positive $x$-axis.

---

### Finding the Resultant of the First Two Displacements

Before we can cancel out the motion, we must find the combined effect of the first two displacements. Let's find the resultant phasor $\vec{A}_{12}$ of $\vec{A}_1$ and $\vec{A}_2$.

Using the vector addition formula, the magnitude of the resultant is:

$$
A_{12} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \Delta \phi}
$$

Substituting $A_1 = A$, $A_2 = A$, and the phase difference $\Delta \phi = \frac{2\pi}{3}$:

$$
A_{12} = \sqrt{A^2 + A^2 + 2A^2 \cos\left(\frac{2\pi}{3}\right)}
$$

Since $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$, we get:

$$
A_{12} = \sqrt{A^2 + A^2 + 2A^2 \left(-\frac{1}{2}\right)} = \sqrt{A^2 + A^2 - A^2} = A
$$

This is a beautiful geometric result!

When two vectors of equal magnitude $A$ have an angle of $120^\circ$ between them, their resultant also has a magnitude of exactly $A$.

Now, let's find the direction of this resultant.

Since the two vectors are equal in magnitude, their resultant must lie exactly along the angle bisector.

The angle bisector of $120^\circ$ is $60^\circ$, or $\frac{\pi}{3}$ radians.

So, the combined displacement of the first two waves is:

$$
x_{12}(t) = A \sin\left(\omega t + \frac{\pi}{3}\right)
$$

---

### Canceling the Motion with the Third Displacement

To bring the mass to complete rest, the third displacement $x_3(t) = B \sin(\omega t + \phi)$ must completely cancel out the resultant displacement $x_{12}(t)$:

$$
x_3(t) + x_{12}(t) = 0 \implies x_3(t) = -x_{12}(t)
$$

In the phasor world, this means the third phasor $\vec{A}_3$ must be exactly equal in magnitude and opposite in direction to the resultant phasor $\vec{A}_{12}$.

1. **Magnitude ($B$):** 
   $$
   B = A_{12} = A
   $$

2. **Phase Angle ($_\phi$):** 
   To point in the exact opposite direction of $\frac{\pi}{3}$, we must add $\pi$ ($180^\circ$) to the phase angle:
   $$
   \phi = \frac{\pi}{3} + \pi = \frac{4\pi}{3}
   $$

Thus, the third displacement must have an amplitude of **$A$** and a phase angle of **$\frac{4\pi}{3}$**.

This perfectly matches **Option (b)**.

Select Answer:

Visualized Solution

Visualizing the Displacements as Phasors

The Condition for Complete Rest

Resultant of $x_{1} (t)$ and $x_{2} (t)$

Substituting the Amplitudes and Phase

Calculating $A_{12}$

Determining the Resultant Phase $θ_{12}$

Balancing Condition

Determining Amplitude $B$

Determining Phase Angle $ϕ$

Matching with the Options

Alternative Analytical Approach

The Sigma Insight: Simple Harmonic Motion (SHM)

Introduction to Superposition and Phasors

The Magic of Phasors

Finding the Resultant of the First Two Displacements

Canceling the Motion with the Third Displacement

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