Animated Solution for Physics - Oscillations: Two simple harmonic motions are represented by the equations $y_1=10\sin \left(3\pi t+\frac{\pi }{4}\right)$ and $y_2=5(\sin 3\pi t+\sqrt{3}\cos 3\pi t)$. Their amplitudes are in the ratio of \_\_\_\_\_\_.

Simple Harmonic Motion (SHM) is one of the most fundamental concepts in physics, describing everything from the swing of a pendulum to the vibration of atoms in a crystal lattice.

Often, a physical system is subjected to multiple harmonic influences simultaneously.

When this happens, we use the Principle of Superposition, which states that the net displacement is the algebraic sum of the individual displacements.

In this problem, we are given two simple harmonic motions represented by the equations:

Our goal is to find the ratio of their amplitudes, $A_1:A_2$.

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The first equation is already presented in the standard form of a simple harmonic wave:

By direct comparison with $y_1=10\sin \left(3\pi t+\frac{\pi }{4}\right)$, we can immediately identify the key parameters of the first motion: - Amplitude ($A_1$): $10$ units - Angular Frequency (${\omega }_1$): $3\pi$ rad/s - Initial Phase (${\varphi }_1$): $\frac{\pi }{4}$ rad

Thus, we have our first amplitude:

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The second equation, $y_2=5(\sin 3\pi t+\sqrt{3}\cos 3\pi t)$, is a linear combination of a sine and a cosine function of the same frequency.

To find its net amplitude, we must combine these two terms into a single sinusoidal expression.

Let's first expand the expression:

This is of the form:

where $a=5$ and $b=5\sqrt{3}$.

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An elegant way to combine sine and cosine functions of the same frequency is by using phasors (vector representation of harmonic quantities).

Since a cosine wave leads a sine wave by a phase difference of $\frac{\pi }{2}$ radians ($90^{\circ }$), we can represent them as two perpendicular vectors: 1. A horizontal vector of magnitude $a=5$ representing the sine component. 2. A vertical vector of magnitude $b=5\sqrt{3}$ representing the cosine component.

The resultant vector represents the combined wave, and its length is the resultant amplitude $A_2$.

Using the Pythagorean theorem for perpendicular vectors:

Substituting our values:

Thus, the amplitude of the second SHM is also $10$ units.

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To write the complete equation for $y_2$, we can also find the resultant phase angle $\varphi$:

Therefore, the combined equation for $y_2$ is:

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Now that we have both amplitudes: - $A_1=10$ - $A_2=10$

The ratio of their amplitudes is:

Thus, the ratio of their amplitudes is $1:1$.