Three simple harmonic motions in the same direction having the same amplitude a and same period are

Q: Why can we represent simple harmonic motions as vectors (phasors)?

Since simple harmonic motion can be projected as uniform circular motion, the displacement can be represented as the projection of a rotating vector (phasor). Adding these vectors geometrically is equivalent to adding the trigonometric functions, which often simplifies the math significantly.

Q: Why did we group $y_1$ and $y_3$ first instead of adding them sequentially?

Grouping $y_1$ (phase $0^\circ$) and $y_3$ (phase $90^\circ$) takes advantage of symmetry. Their average phase is exactly $45^\circ$, which matches the phase of $y_2$. This alignment allows us to add the remaining term directly without complex phase calculations.

Q: Is the resultant of any number of superposed SHMs always simple harmonic?

Yes, as long as all the individual SHMs have the same frequency and are along the same line, their superposition will always result in a single simple harmonic motion of the same frequency.

Q: Why is the energy of the resultant motion not simply the sum of the individual energies?

This is due to interference! When waves are in phase or partially in phase, they interfere constructively, causing the resultant amplitude to be larger than the individual amplitudes. Since energy is proportional to the square of the amplitude ($E \\propto A^2$), the total energy is greater than the sum of individual energies ($E_R > 3E$).

Q: What would happen if the phase difference between successive SHMs was $120^\circ$ instead of $45^\circ$?

If three identical SHMs differ in phase by $120^\circ$ successively (i.e., $0^\circ$, $120^\circ$, $240^\circ$), their phasors form a closed equilateral triangle. In this case, the resultant amplitude would be exactly zero, representing complete destructive interference!

Question

Three simple harmonic motions in the same direction having the same amplitude $a$ and same period are superposed. If each differ in phase from the next by $45^\circ$, then

Options:

(A) the resultant amplitude is $(1+\sqrt{2})a$

(B) the phase of the resultant motion relative to the first is $90^\circ$

(C) the energy associated with the resulting motion is $(3+2\sqrt{2})$ times the energy associated with any single motion

(D) the resulting motion is not simple harmonic

Sigma Percentile · Accepted Answer

### The Symphony of Superposition

Imagine standing on a quiet beach, watching waves roll in from the ocean.

When two or more waves meet, they do not crash and destroy each other; instead, they pass through one another, temporarily combining their strengths in a beautiful dance governed by the **Principle of Superposition**.

In this problem, we are exploring the superposition of three identical simple harmonic motions (SHMs) traveling in the same direction.

Each wave has the same amplitude $a$ and the same time period $T$ (and thus the same angular frequency $\omega$), but they are slightly out of step with one another—each differing in phase from the next by exactly $45^\circ$.

Our goal is to find the resultant amplitude, phase, and energy of this combined motion and see which of the given options hold true.

### The Setup: Visualizing the Waves

Let's write down the mathematical equations for our three individual simple harmonic motions.

Since they all have the same frequency $\omega$ and amplitude $a$, we can choose our reference phase such that the first motion starts at zero phase:

$y_1 = a \sin(\omega t)$

Since each wave differs from the next by $45^\circ$, the second and third motions are given by:

$y_2 = a \sin(\omega t + 45^\circ)$

$y_3 = a \sin(\omega t + 90^\circ)$

According to the principle of superposition, the net displacement $y$ at any instant is simply the algebraic sum of these three individual displacements:

$y = y_1 + y_2 + y_3$

### The Phasor Shortcut: Geometry to the Rescue

Before diving into heavy trigonometry, let's look at this problem through the lens of **Phasors**.

A phasor is a rotating vector whose length represents the amplitude of the SHM, and whose angle with the horizontal axis represents its phase.

Adding simple harmonic motions of the same frequency is mathematically identical to adding their corresponding phasor vectors!

Let's represent our three motions as vectors:
- $\vec{A}_1$ has length $a$ and points along the positive x-axis ($0^\circ$).
- $\vec{A}_2$ has length $a$ and points at an angle of $45^\circ$.
- $\vec{A}_3$ has length $a$ and points along the positive y-axis ($90^\circ$).

Notice the beautiful symmetry here!

The vectors $\vec{A}_1$ and $\vec{A}_3$ are perpendicular to each other.

Their resultant, let's call it $\vec{A}_{13}$, will lie exactly along the angle bisector—which is $45^\circ$!

The magnitude of this resultant is given by the Pythagorean theorem:

$A_{13} = \sqrt{a^2 + a^2} = a\sqrt{2}$

Now, we need to add the second vector $\vec{A}_2$ to this intermediate resultant.

Since $\vec{A}_2$ also points exactly along $45^\circ$, the two vectors $\vec{A}_{13}$ and $\vec{A}_2$ are **collinear**!

When two vectors point in the exact same direction, we can find their resultant magnitude by simply adding their lengths directly:

$A_R = A_{13} + A_2 = a\sqrt{2} + a = a(1 + \sqrt{2})$

And because both vectors point along $45^\circ$, the final resultant vector $\vec{A}_R$ also points at $45^\circ$.

This geometric insight is incredibly elegant! It tells us instantly that:
1. The resultant amplitude is $a(1 + \sqrt{2})$.
2. The phase of the resultant motion relative to the first motion ($y_1$) is $45^\circ$.

### The Analytical Path: Trigonometric Elegance

Let's verify this beautiful geometric result using pure algebra to ensure absolute mathematical rigor.

We want to find the sum:

$y = a \sin(\omega t) + a \sin(\omega t + 45^\circ) + a \sin(\omega t + 90^\circ)$

Using our symmetry trick, let's group the first and third terms together first:

$y_1 + y_3 = a [\sin(\omega t) + \sin(\omega t + 90^\circ)]$

We can apply the sum-to-product trigonometric identity:

$\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$

Setting $C = \omega t + 90^\circ$ and $D = \omega t$, we get:

$y_1 + y_3 = 2a \sin\left(\frac{2\omega t + 90^\circ}{2}\right) \cos\left(\frac{90^\circ}{2}\right)$

$y_1 + y_3 = 2a \sin(\omega t + 45^\circ) \cos(45^\circ)$

Since $\cos(45^\circ) = \frac{1}{\sqrt{2}}$, this simplifies to:

$y_1 + y_3 = 2a \sin(\omega t + 45^\circ) \left(\frac{1}{\sqrt{2}}\right) = \sqrt{2} a \sin(\omega t + 45^\circ)$

Now, we add the second term $y_2$ back into our equation:

$y = (y_1 + y_3) + y_2$

$y = \sqrt{2} a \sin(\omega t + 45^\circ) + a \sin(\omega t + 45^\circ)$

Factoring out the common sine term, we get:

$y = (1 + \sqrt{2}) a \sin(\omega t + 45^\circ)$

This is a perfect match with our phasor result!

The resultant motion is indeed a simple harmonic motion with an amplitude of $A_R = (1 + \sqrt{2})a$ and a phase shift of $45^\circ$ relative to the first motion.

Therefore, **Option (a) is correct**, and **Option (b) is incorrect** (since the phase is $45^\circ$, not $90^\circ$).

### Energy: The Power of Amplitude

Now, let's look at the energy associated with this resultant motion, which is the subject of option (c).

The total mechanical energy $E$ of a simple harmonic oscillator is directly proportional to the square of its amplitude:

$E \propto A^2$

For a single motion of amplitude $a$, the energy is:

$E_{\text{single}} = k a^2$

For our resultant motion of amplitude $A_R = (1 + \sqrt{2})a$, the energy is:

$E_{\text{resultant}} = k A_R^2 = k [a(1 + \sqrt{2})]^2$

Let's expand the squared term:

$(1 + \sqrt{2})^2 = 1 + 2 + 2\sqrt{2} = 3 + 2\sqrt{2}$

Substituting this back, we find:

$E_{\text{resultant}} = (3 + 2\sqrt{2}) k a^2 = (3 + 2\sqrt{2}) E_{\text{single}}$

This means the energy of the resultant motion is exactly $(3 + 2\sqrt{2})$ times the energy of any single motion!

Therefore, **Option (c) is correct**.

Finally, since the resultant equation $y = (1 + \sqrt{2}) a \sin(\omega t + 45^\circ)$ is a single sinusoidal function of time, the resulting motion is definitely simple harmonic.

Thus, **Option (d) is incorrect**.

### Conclusion: The Harmony of Math and Physics

By combining the physical principle of superposition with the elegant geometry of phasors and trigonometric identities, we have successfully solved this classic JEE problem!

The correct options are **(a)** and **(c)**.

Animated Solution for Physics - Oscillations: Three simple harmonic motions in the same direction having the same amplitude $a$ and same period are superposed. If each differ in phase from the next by $4 5^{\circ}$ , then

Select Answer:

Visualized Solution

Representing SHMs as Phasors

Principle of Superposition

Grouping Symmetric Terms

Applying Trigonometric Identity

Resultant of $y_{1}$ and $y_{3}$

Simplifying $y_{1} + y_{3}$

Total Resultant Displacement

Resultant Parameters

Energy of Resultant Motion

Simplifying the Energy Ratio

Final Verdict

The Sigma Insight: Simple Harmonic Motion (SHM)

The Symphony of Superposition

The Setup

The Phasor Shortcut

The Analytical Path

Energy

Conclusion

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