Two simple harmonic motions are represented by the equations x1 = 5 sin(2πt + π/4) and x2 = 5√2(sin

Q: Why can we treat sine and cosine terms as perpendicular vectors?

Sine and cosine functions of the same frequency have a constant phase difference of $\frac{\pi}{2}$ (or $90^\circ$). In phasor diagrams, phase difference corresponds to the geometric angle between vectors, making them perfectly perpendicular.

Q: Why can't we just add the coefficients $5\sqrt{2} + 5\sqrt{2}$ to get the total amplitude?

Because the two waves are not in phase! They don't reach their maximum values at the same time. You must add them vectorially using the Pythagorean theorem, not algebraically.

Q: What if the frequencies of the sine and cosine terms were different?

If the frequencies were different (e.g., $\sin 2\pi t$ and $\cos 3\pi t$), the resulting motion would not be a Simple Harmonic Motion. It would be a complex periodic wave, and it wouldn't have a single, constant amplitude.

Q: Is there a purely trigonometric way to solve this without using vectors?

Absolutely! You can multiply and divide the expression by $\sqrt{a^2 + b^2}$ (which is 10 here). This transforms the equation into $10 \left( \frac{1}{\sqrt{2}} \sin 2\pi t + \frac{1}{\sqrt{2}} \cos 2\pi t \right)$, which simplifies to $10 \sin\left(2\pi t + \frac{\pi}{4}\right)$ using the $\sin(A+B)$ identity.

Q: Does the initial phase angle $\frac{\pi}{4}$ in the first equation affect its amplitude?

Not at all. The phase angle only determines the starting position of the particle at $t=0$. It shifts the wave horizontally in time but does not stretch or shrink its vertical height (amplitude).

Question

Two simple harmonic motions are represented by the equations
$x_1 = 5 \sin \left( 2\pi t + \frac{\pi}{4} \right)$ and
$x_2 = 5\sqrt{2} (\sin 2\pi t + \cos 2\pi t)$
The amplitude of second motion is .............. times the amplitude in first motion.

Sigma Percentile · Accepted Answer

### The Magic of Phasors

Simple Harmonic Motion (SHM) is one of the most elegant concepts in physics. While we often write it as a trigonometric equation, its true beauty is revealed when we visualize it as a rotating vector, known as a **phasor**.

Imagine a vector rotating in a circle at a constant angular velocity $\omega$. The projection of this vector on the y-axis (or x-axis) perfectly traces out a sine (or cosine) wave over time. This geometric perspective turns complex trigonometric additions into simple vector geometry!

### Decoding the First Motion

Let's look at our first equation:

$$x_1 = 5 \sin \left( 2\pi t + \frac{\pi}{4} ight)$$

This is already in the pristine, standard form of an SHM, which is $x = A \sin(\omega t + \phi)$. By simply comparing the two, we can instantly extract the amplitude. The coefficient in front of the sine function is the maximum displacement from the mean position.

Therefore, the amplitude of the first motion is:

$$A_1 = 5$$

### Superposition: When Sine Meets Cosine

Now, let's tackle the second equation, which looks a bit more intimidating:

$$x_2 = 5\sqrt{2} (\sin 2\pi t + \cos 2\pi t)$$

If we expand this, we get:

$$x_2 = 5\sqrt{2} \sin 2\pi t + 5\sqrt{2} \cos 2\pi t$$

This represents the **superposition** of two separate simple harmonic motions. Notice that they share the exact same angular frequency ($2\pi$), but one is a sine wave and the other is a cosine wave.

Mathematically, a cosine wave is just a sine wave shifted by $90^\circ$ (or $\frac{\pi}{2}$ radians). In our phasor diagram, this means the sine term can be drawn as a vector along the horizontal axis, and the cosine term as a vector along the vertical axis. They are perfectly **perpendicular** to each other!

### The Vector Addition

To find the resultant amplitude $A_2$, we don't need to memorize complex trigonometric identities. We just need to find the vector sum of these two perpendicular phasors.

For any expression of the form $a \sin 	heta + b \cos 	heta$, the resultant amplitude $R$ is simply the hypotenuse of the right-angled triangle formed by $a$ and $b$:

$$R = \sqrt{a^2 + b^2}$$

In our case, both $a$ and $b$ are equal to $5\sqrt{2}$. Let's plug them into our Pythagorean formula:

$$A_2 = \sqrt{(5\sqrt{2})^2 + (5\sqrt{2})^2}$$

$$A_2 = \sqrt{50 + 50}$$

$$A_2 = \sqrt{100} = 10$$

The resultant amplitude of the second motion is exactly 10.

### The Final Verdict

The question asks for the ratio of the amplitude of the second motion to the amplitude of the first motion.

$$	ext{Ratio} = \frac{A_2}{A_1} = \frac{10}{5} = 2$$

Thus, the amplitude of the second motion is **2 times** the amplitude of the first motion.

*Bonus Insight:* If you calculate the phase angle of the second motion using $	an \phi = \frac{b}{a} = \frac{5\sqrt{2}}{5\sqrt{2}} = 1$, you get $\phi = \frac{\pi}{4}$. This means $x_2$ can be rewritten as $10 \sin\left(2\pi t + \frac{\pi}{4}ight)$. It is perfectly in phase with $x_1$, just twice as large!

Animated Solution for Physics - Oscillations: Two simple harmonic motions are represented by the equations $x_{1} = 5 sin (2 π t + \frac{π}{4})$ and $x_{2} = 52 (sin 2 π t + cos 2 π t)$ The amplitude of second motion is .............. times the amplitude in first motion.

Enter Numerical Value:

Visualized Solution

$x_{1}$ and $x_{2}$ Representation

Standard SHM Form

Amplitude of First Motion

Superposition in Second Motion

Resultant Amplitude Formula

Calculating $A_{2}$

Ratio of Amplitudes

Phase Angle Insight

The Sigma Insight: Simple Harmonic Motion (SHM)

The Magic of Phasors

Decoding the First Motion

Superposition

The Vector Addition

The Final Verdict

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