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If the equation for displacement of two particles executing S.H.M. is given by \(\mathrm{y}_{1}=2 \sin (10 \mathrm{t}+\theta)\) and \(\mathrm{y}_{2}=3 \cos 10 \mathrm{t}\) respectively, then the phase difference between the velocity of two particles will be \(\ldots \ldots \ldots\) (A) \(-\theta\) (B) \(\theta\) (C) \(\theta-(\pi / 2)\) (D) \(\theta+(\pi / 2)\).

Short Answer

Expert verified
The phase difference between the velocity of two particles will be \(\theta - (\frac{\pi}{2})\).

Step by step solution

01

Find the velocities of the particles#

To find the velocities of the particles, we will differentiate the displacement equations with respect to time: Particle 1: The displacement equation is given by \(y_1 = 2 \sin(10t + \theta)\) Taking the derivative of \(y_1\) with respect to time \(t\) gives the velocity: \(v_1 = \frac{dy_1}{dt} = 2 \cdot 10 \cos(10t + \theta) = 20 \cos(10t + \theta)\) Particle 2: The displacement equation is given by \(y_2 = 3 \cos(10t)\) Taking the derivative of \(y_2\) with respect to time \(t\) gives the velocity: \(v_2 = \frac{dy_2}{dt} = -3 \cdot 10 \sin(10t) = -30 \sin(10t)\)
02

Calculate the phase difference between the two velocities#

Now, we want to calculate the phase difference between these two velocities. Let's rewrite the velocities as a product of a trigonometric function and a constant to express it as a simple harmonic equation: \(v_1 = 20 \cos(10t + \theta) = 20 \cos (\omega t + \phi_1)\) with \(\omega = 10\) and \(\phi_1 = \theta\) \(v_2 = -30 \sin(10t) = -30 \cos \left(10t - \frac{\pi}{2}\right) = 30 \cos (\omega t + \phi_2)\) with \(\omega = 10\) and \(\phi_2 = -\frac{\pi}{2}\) The phase difference between the two velocities can be calculated by subtracting \(\phi_1\) from \(\phi_2\): \(\Delta \phi = \phi_2 - \phi_1 = (-\frac{\pi}{2}) - \theta\) Therefore, the phase difference between the velocity of two particles is: \(\Delta \phi = \theta - (\frac{\pi}{2})\) The correct answer is (C).

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Key Concepts

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

Phase Difference
When analyzing simple harmonic motion (SHM), the term 'phase difference' refers to how much one wave or motion is shifted in relation to another. In essence, it's the angular difference between the initial stages of two periodic processes.
For example, in two harmonic motions represented by sine and cosine functions, like the ones given in the problem, the phase difference plays a critical role in determining how these motions interact or align with each other. Specifically, for SHM characterized by equations such as \(y_1 = 2 \sin(10t + \theta)\) and \(y_2 = 3 \cos(10t)\), the phase difference becomes apparent when calculated from the velocities.
  • The given velocities are expressed as \(v_1 = 20 \cos(10t + \theta)\) and \(v_2 = -30 \sin(10t) = -30 \cos \left(10t - \frac{\pi}{2}\right)\).
  • Thus, finding the phase difference between these two velocities is necessary to understand their dynamic interaction.
As determined in the solution, this phase difference is \(\theta - \frac{\pi}{2}\), representing how far one wave leads or lags behind the other.
This concept is significant in physics as it explains scenarios ranging from sound waves to alternating currents.
Velocity of Particles
In dynamics involving simple harmonic motion, the velocity of a particle describes how quickly it changes its position over time. Velocity in SHM is derived by differentiating the displacement equation with respect to time.
Here's how it's done for the particles in this exercise:
  • For Particle 1, whose displacement is given by \(y_1 = 2 \sin(10t + \theta)\), the velocity is derived as \(v_1 = \frac{dy_1}{dt} = 20 \cos(10t + \theta)\).
  • For Particle 2, with displacement \(y_2 = 3 \cos(10t)\), its velocity becomes \(v_2 = \frac{dy_2}{dt} = -30 \sin(10t)\).
The calculation involves using trigonometric identities and differentiation rules encountered in calculus.
This forms the basis for analysis and comparison of their motions and leads directly to understanding the phase difference. In SHM, recognizing how velocity varies with time is crucial to predicting and explaining the behavior of oscillating systems.
This technique is pivotal in many areas, such as predicting tides, analyzing bridges' vibrations, and tuning musical instruments, all of which depend heavily on precise calculations of velocity in SHM.
Trigonometric Functions
Trigonometric functions, including sine and cosine, are central in describing simple harmonic motion. These functions model the repeating oscillations observed in systems, allowing us to predict their behavior under various conditions.
For instance, in the mathematical models for the displacements given in the problem, these functions let's express:
  • For Particle 1, the equation \(y_1 = 2 \sin(10t + \theta)\) employs a sine function to describe its periodic motion.
  • For Particle 2, the cosine function is used in \(y_2 = 3 \cos(10t)\).
These functions relate the displacement of a particle to its phase angle, a combination of time, frequency, and initial phase.

It's important to know that:
  • The derivative of sine is cosine, and the derivative of cosine is negative sine, assisting in finding velocities from displacements in SHM.
  • Trigonometric identities also aid in transforming and simplifying expressions, like converting between sine and cosine to determine phase differences.
In physics, these periodic functions not only provide insight into SHM but also underpin the study of waves, signal processing, and many other applications in engineering and sciences.

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Most popular questions from this chapter

As shown in figure, two light springs having force constants \(\mathrm{k}_{1}=1.8 \mathrm{~N} \mathrm{~m}^{-1}\) and \(\mathrm{k}_{2}=3.2 \mathrm{~N} \mathrm{~m}^{-1}\) and a block having mass \(\mathrm{m}=200 \mathrm{~g}\) are placed on a frictionless horizontal surface. One end of both springs are attached to rigid supports. The distance between the free ends of the spring is \(60 \mathrm{~cm}\) and the block is moving in this gap with a speed \(\mathrm{v}=120 \mathrm{~cm} \mathrm{~s}^{-1}\).When the block is moving towards \(k_{1}\), what will be the time taken for it to get maximum compressed from point \(\mathrm{C}\) ? (A) \(\pi \mathrm{s}\) (B) \((2 / 3) \mathrm{s}\) (C) \((\pi / 3) \mathrm{s}\) (D) \((\pi / 4) \mathrm{s}\)

If two SHM's are given by the equation \(\mathrm{y}_{1}=0.1 \sin [\pi \mathrm{t}+(\pi / 3)]\) and \(\mathrm{y}_{2}=0.1 \cos \pi \mathrm{t}\), then the phase difference between the velocity of particle 1 and 2 is \(\ldots \ldots \ldots\) (A) \(\pi / 6\) (B) \(-\pi / 3\) (C) \(\pi / 3\) (D) \(-\pi / 6\)

A simple pendulum having length \(\ell\) is given a small angular displacement at time \(t=0\) and released. After time \(t\), the linear displacement of the bob of the pendulum is given by \(\ldots \ldots \ldots \ldots\) (A) \(x=a \sin 2 p \sqrt{(\ell / g) t}\) (B) \(\mathrm{x}=\mathrm{a} \cos 2 \mathrm{p} \sqrt{(\mathrm{g} / \ell) t}\) (C) \(\mathrm{x}=\mathrm{a} \sin \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\) (D) \(\mathrm{x}=\mathrm{a} \cos \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\)

When an elastic spring is given a displacement of \(10 \mathrm{~mm}\), it gains an potential energy equal to \(\mathrm{U}\). If this spring is given an additional displacement of \(10 \mathrm{~mm}\), then its potential energy will be.............. (A) \(\mathrm{U}\) (B) \(2 \mathrm{U}\) (C) \(4 \mathrm{U}\) (D) \(\mathrm{U} / 4\).

A spring is attached to the center of a frictionless horizontal turn table and at the other end a body of mass \(2 \mathrm{~kg}\) is attached. The length of the spring is \(35 \mathrm{~cm}\). Now when the turn table is rotated with an angular speed of \(10 \mathrm{rad} \mathrm{s}^{-1}\), the length of the spring becomes \(40 \mathrm{~cm}\) then the force constant of the spring is \(\ldots \ldots \mathrm{N} / \mathrm{m}\). (A) \(1.2 \times 10^{3}\) (B) \(1.6 \times 10^{3}\) (C) \(2.2 \times 10^{3}\) (D) \(2.6 \times 10^{3}\)

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