/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 A particle moves in \(x y\)-plan... [FREE SOLUTION] | 91Ó°ÊÓ

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A particle moves in \(x y\)-plane according to the rule \(x=a \sin \omega t\) and \(y=a \cos \omega t\). The particle follows (a) an elliptical path (b) a circular path (c) a parabolic path (d) a straight line path inclined equally to \(x\) and \(y\)-axis

Short Answer

Expert verified
The particle follows a circular path.

Step by step solution

01

Understanding the Problem

Identify the equations of motion given for the particle. We have two parametric equations describing the motion: \(x = a \sin \omega t\) and \(y = a \cos \omega t\). These equations describe how the x and y coordinates change over time.
02

Recall Trigonometric Identity

Recall the trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\). We will utilize this identity to eliminate the parameter \(\omega t\) from the given equations.
03

Substitute and Simplify

Substitute \(x = a \sin \omega t\) and \(y = a \cos \omega t\) into the identity. We obtain: \(\left(\frac{x}{a}\right)^2 + \left(\frac{y}{a}\right)^2 = \sin^2 \omega t + \cos^2 \omega t = 1\).
04

Recognize the Path Form

The resulting equation \(\frac{x^2}{a^2} + \frac{y^2}{a^2} = 1\) is the standard form of the equation of a circle, centered at the origin with radius \(a\).
05

Conclusion

The path described by the equations \(x = a \sin \omega t\) and \(y = a \cos \omega t\) is a circle with radius \(a\). Therefore, the particle follows a circular path.

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Key Concepts

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

Circular Motion
In physics, circular motion is a type of motion whereby an object moves along the circumference of a circle or rotation along a circular path. **The given parametric equations** in the problem describe such a motion where the coordinates are functions of time. This is crucial for understanding how motion is modeled. The equations provided—\(x = a \sin \omega t\) and \(y = a \cos \omega t\)—specify the location of a particle in the \(x y\)-plane as it moves over time.
Typically, in circular motion:
  • The radius of the circle, \(a\), remains constant.
  • The angle \(\omega t\) gives us information about the particle's position on the circle at a given time.
  • \(\omega\) is the angular velocity, representing how fast the particle travels around the circle.

As shown in the solution, by applying the trigonometric identity, we confirm that the particle follows a circular path with radius \(a\), making it a crucial concept in analyzing such motions.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for all angles. They are very useful in simplifying and solving parametric equations. In this exercise, the identity \(\sin^2 \theta + \cos^2 \theta = 1\) was utilized to eliminate the parameter \(\omega t\) and reveal the nature of the path.
Let's break this down:
  • This identity allows us to combine the trigonometric components of the equations \(x = a \sin \omega t\) and \(y = a \cos \omega t\).
  • By rewriting the components as fractions, \(\left(\frac{x}{a}\right)^2 + \left(\frac{y}{a}\right)^2 = 1\), we formed a recognizable pattern, which corresponds to a circle.
  • Using identities simplifies the system, connecting complex equations to known geometric shapes.

Recognizing these identities helps in solving problems efficiently, especially in motion, determining the true path or trajectory when visualizing physical movements.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, involves using a coordinate system to explore geometric shapes and their properties. In this case, it plays a crucial role in transforming the parametric equations of motion into a geometric form we can recognize.
When examining the equations \(x = a \sin \omega t\) and \(y = a \cos \omega t\), coordinate geometry helps us:
  • Relate the equations to the shape in the coordinate plane—here, a circle.
  • Translate motion equations into an equation understood visually and algebraically: the circle equation \(\frac{x^2}{a^2} + \frac{y^2}{a^2} = 1\).
  • Identify properties such as center and radius from the geometric form, with the center at the origin and radius \(a\).

Unlike basic graphs showing position over time, coordinate geometry allows us to see motion paths and geometric shapes, illustrating deeper connections in physics and mathematics.

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Most popular questions from this chapter

A mass \(M\), attached to a spring, oscillates with a period of 2 s. If the mass is increased by \(4 \mathrm{~kg}\), the time period increases by \(1 \mathrm{~s}\). Assuming that Hooke's law is obeyed, the initial mass \(M\) was (a) \(3.2 \mathrm{~kg}\) (b) \(1 \mathrm{~kg}\) (c) \(2 \mathrm{~kg}\) (d) \(8 \mathrm{~kg}\)

The amplitude of SHM \(y=2(\sin 5 \pi t+\sqrt{3} \cos 5 \pi t)\) is \(\quad\) (a) 2 (b) \(2 \sqrt{2}\) (c) 4 (d) \(2 \sqrt{3}\)

Two pendulums of length \(1 \mathrm{~m}\) and \(16 \mathrm{~m}\) start vibrating one behind the other from the same stand. At some instant, the two are in the mean position in the same phase. The time period of shorter pendulum is \(T\). The minimum time after which the two threads of the pendulum will be one behind the other is (a) \(T / 4\) (b) \(T / 3\) \(\begin{array}{ll}\text { (c) } 4 T / 3 & \text { (d) } 4 T\end{array}\)

Assertion The amplitude of a particle executing SHM with a frequency of \(60 \mathrm{~Hz}\) is \(0.01 \mathrm{~m}\). The maximum value of acceleration of the particle is \(\pm 144 \pi^{2} \mathrm{~ms}^{-2}\). Reason Acceleration amplitude \(=\omega^{2} A\),where \(A\) is displacement amplitude.

A heavy sphere of mass \(m\) is suspended by string of length \(l\). The sphere is made to revolve about a vertical line passing through the point of suspension in a horizontal circle such that the string always remains inclined to the vertical at an angle \(\theta\). What is its period of revolution? (a) \(T=2 \pi \sqrt{\frac{l}{g}}\) (b) \(T=2 \pi \sqrt{\frac{l \cos \theta}{g}}\) (c) \(T=2 \pi \sqrt{\frac{l \sin \theta}{g}}\) (d) \(T=2 \pi \sqrt{\frac{l \tan \theta}{g}}\)

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