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Two simple harmonic motions are represented by the equations \(y_{1}=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)\) and \(y_{2}=0.1 \cos \pi t\). The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is (a) \(\frac{\pi}{6}\) (b) \(\frac{-\pi}{3}\) (c) \(\frac{\pi}{3}\) (d) \(\frac{-\pi}{6}\)

Short Answer

Expert verified
The phase difference is \( \frac{-\pi}{6} \), corresponding to option (d).

Step by step solution

01

Find the Velocity of Particle 1

The velocity of a particle given by simple harmonic motion can be found by differentiating the position function with respect to time, i.e., \( v_1 = \frac{dy_1}{dt} \). For particle 1, \( y_1=0.1 \sin (100 \pi t+\frac{\pi}{3}) \).Differentiate to find \( v_1 \):\[ v_1 = \frac{d}{dt} [0.1 \sin (100 \pi t+\frac{\pi}{3})] = 0.1 \times 100 \pi \cos (100 \pi t+\frac{\pi}{3}) \]Thus, \( v_1 = 10 \pi \cos (100 \pi t+\frac{\pi}{3}) \).
02

Find the Velocity of Particle 2

For particle 2, the position function is \( y_2 = 0.1 \cos \pi t \).To find the velocity, differentiate with respect to time:\[ v_2 = \frac{d}{dt} [0.1 \cos \pi t] = -0.1 \pi \sin \pi t \]Thus, \( v_2 = -0.1 \pi \sin \pi t \).
03

Write the Velocity Equations in Standard Form

For trigonometric simplification, let's express the velocity equations in a standard form so that phase comparison is easier.\[ v_1 = 10 \pi \cos (100 \pi t+\frac{\pi}{3}) \]\[ v_2 = -0.1 \pi \sin (\pi t) \] can be rewritten as:\[ v_2 = 0.1 \pi \cos (\pi t + \frac{\pi}{2}) \]This is using the identity \( \sin A = \cos (A - \frac{\pi}{2}) \).
04

Calculate the Phase Difference

Now calculate the phase difference by comparing the arguments of the cosine functions from the standardized velocity equations:For \( v_1 \), the phase is \( 100\pi t + \frac{\pi}{3} \).For \( v_2 \), the phase is \( \pi t + \frac{\pi}{2} \).The phase difference is:\[ (100\pi t + \frac{\pi}{3}) - (\pi t + \frac{\pi}{2}) = 99\pi t + \frac{\pi}{3} - \frac{\pi}{2} \]Simplifying gives:\[ \frac{\pi}{3} - \frac{\pi}{2} = \frac{-\pi}{6} \]
05

Determine the Correct Answer Option

The calculated phase difference is \( \frac{-\pi}{6} \), which corresponds to option (d).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Phase Difference
In simple harmonic motion, phase difference refers to the difference in phase between two oscillating systems. Phase difference tells us how two waves or harmonic motions are related in terms of their oscillation cycle. This is a crucial concept because it helps in determining how in-sync or out-of-sync two wave patterns are.

Imagine two runners on a circular track. Even if they start at the same point, if one is ahead or behind, this difference is akin to phase difference. In this exercise, both motions are expressed with trigonometric functions, sine and cosine, and each has its argument, depicting its phase.

Once velocities are found, the phase difference between the two systems is calculated by comparing the arguments within the trigonometric expressions received after differentiating the original position functions. The phase difference is not fixed; it depends on time and the characteristics of the oscillations. But for particles 1 and 2, the relevant values were determined to evaluate to a constant phase difference of \( \frac{-\pi}{6} \).
Velocity Calculation in Simple Harmonic Motion
In the realm of simple harmonic motion, the velocity of a particle can be derived from its position function by taking the derivative with respect to time. This is because velocity is the rate of change of displacement with time.

Let's consider particle 1, with position given by \( y_1 = 0.1 \sin(100\pi t + \frac{\pi}{3}) \). By differentiating with respect to \( t \), we find its velocity: \( v_1 = 10\pi \cos(100\pi t + \frac{\pi}{3}) \). The differentiation involves applying the chain rule, capturing how quickly the positional change converts into velocity across the time axis.

For particle 2, expressing its position as \( y_2 = 0.1 \cos(\pi t) \), the velocity is determined by \( v_2 = -0.1\pi \sin(\pi t) \). Here, the negative sign indicates a phase inversion, common in trigonometric differentiations. These formulas resulting from derivatives are vital as they depict the velocity forms that aid in subsequent trigonometric simplification and other analyses.
Trigonometric Simplification and Its Impact
Trigonometric simplification often plays a pivotal role in making complex expressions easier to work with, particularly so in simple harmonic motion problems. The key to simplifying is knowing and applying trigonometric identities.

In this exercise, simplifying the velocity function for particle 2 requires expressing it in a standard trigonometric form. Originally \( v_2 = -0.1\pi \sin(\pi t) \) can be re-expressed using the identity \( \sin A = \cos(A - \frac{\pi}{2}) \), which transforms it to \( v_2 = 0.1\pi \cos(\pi t + \frac{\pi}{2}) \). Using these identities makes the function more comparable to \( v_1 = 10\pi \cos(100\pi t + \frac{\pi}{3}) \).

This reformatting is essential for directly comparing phases and identifying phase differences. These identities take potentially intricate trigonometric relationships and transform them into familiar, comparably straightforward expressions. This powerful technique reduces complexity and lays the groundwork for further analysis of harmonic motion's characteristics.

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