/*! 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 150 A particle moves along the \(x\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 150

A particle moves along the \(x\)-axis as per the equation \(x=4+3 \sin (2 \pi t)\). Here \(x\) is in \(\mathrm{cm}\) and \(t\) in seconds. Select the correct alternative(s) (A) The motion of the particle is simple harmonic with mean position at \(x=0\). (B) The motion of the particle is simple harmonic with mean position at \(x=4 \mathrm{~cm}\). (C) The motion of the particle is simple harmonic with mean position at \(x=-4 \mathrm{~cm}\). (D) Amplitude of oscillation is \(3 \mathrm{~cm}\).

Short Answer

Expert verified
The motion of the particle is simple harmonic with mean position at \(x = 4\,\mathrm{cm}\) and amplitude of oscillation is \(3\,\mathrm{cm}\). The correct options are (B) and (D).

Step by step solution

01

Identify Simple Harmonic Motion

A function will represent simple harmonic motion if it takes the form: \(x = A + B\sin(Ct + D)\) or \(x = A + B\cos(Ct + D)\), where A, B, C, and D are constants. Comparing the given equation \(x = 4 + 3\sin(2\pi t)\) to the general form, we can see that it represents simple harmonic motion with A = 4, B = 3, C = 2蟺, and D = 0.
02

Determine Mean Position

The mean position of the particle (equilibrium position) can be found by looking at the constant term in the equation of motion. In our case, it is given by A = 4. Therefore, the motion of the particle is simple harmonic with mean position at \(x = 4\,\mathrm{cm}\).
03

Determine Amplitude

Amplitude of oscillation is the maximum displacement of the particle from its mean position. It is represented by the coefficient of the sinusoidal term in the equation of motion. In our case, it is given by B = 3. Therefore, the amplitude of oscillation is \(3\,\mathrm{cm}\). From our analysis, we can conclude that the correct options are: (B) The motion of the particle is simple harmonic with mean position at \(x = 4\,\mathrm{cm}\). (D) Amplitude of oscillation is \(3\,\mathrm{cm}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Mean Position
In the context of simple harmonic motion, the mean position is a foundational concept. It refers to the equilibrium or central position around which an object oscillates. For the motion described by the equation \(x = 4 + 3 \sin(2\pi t)\), the mean position can be identified directly from the constant term in the equation. The term "4" indicates that the equilibrium position is at \(x = 4 \,\mathrm{cm}\) along the \(x\)-axis.

This equilibrium or mean position is the location to which the particle naturally returns when not subject to any external forces. It's important to note that in simple harmonic motion, movement is symmetrical around this mean position. This means that the particle moves equal distances on either side of this point, allowing us to easily predict its behavior during the oscillatory cycle.
Amplitude of Oscillation
The concept of amplitude is crucial when dealing with oscillatory motion. Amplitude refers to the maximum distance that the oscillating particle travels away from its mean position.

In the expression \(x = 4 + 3 \sin(2\pi t)\), the amplitude is represented by the coefficient of the sinusoidal function, which is "3". Hence, the particle oscillates between \(4 + 3 = 7\,\mathrm{cm}\) and \(4 - 3 = 1\,\mathrm{cm}\). This indicates an amplitude of \(3\,\mathrm{cm}\).

Understanding amplitude helps in predicting the extent of oscillation and allows us to visualize the motion more effectively. A larger amplitude signifies greater energy in the oscillation as the particle moves further from the mean position.
Sinusoidal Function
In simple harmonic motion, sinusoidal functions like \(\sin\theta\) or \(\cos\theta\) play a key role in describing the behavior of oscillating systems.

For the particle's motion described by \(x = 4 + 3 \sin(2\pi t)\), the "sin" term dictates the periodic nature of the motion. The period of this motion, which is the time taken for one complete cycle of oscillation, can be calculated from the angular frequency \(\omega = 2\pi\). The period (T) is given by \(T = \frac{2\pi}{\omega} = 1\,\mathrm{s}\).

The sinusoidal function's shape ensures that the motion is smooth and periodic, creating a wave-like pattern when plotted over time. Sinusoidal functions are therefore fundamental to accurately modeling simple harmonic motions, such as springs, pendulums, and waves, providing insights into both the timing and the path of the oscillating object.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The equation of a wave on a string of linear mass density \(0.04 \mathrm{~kg} \mathrm{~m}^{-1}\) is given by \(y=0.02(m) \sin \left[2 \pi\left(\frac{t}{0.04(s)}-\frac{x}{0.50(m)}\right)\right]\) The tension in the string is (A) \(4.0 \mathrm{~N}\) (B) \(12.5 \mathrm{~N}\) (C) \(0.5 \mathrm{~N}\) (D) \(6.25 \mathrm{~N}\)

A particle of mass \(m\) moves in a straight line. If \(v\) is the velocity at a distance \(x\) from a fixed point on the line and \(v^{2}=a-b x^{2}\), where \(a\) and \(b\) are constant, then (A) The motion continues along the positive \(x\)-direction only. (B) The motion is simple harmonic. (C) The particle oscillates with a frequency equal to \(\frac{\sqrt{b}}{2 \pi}\) (D) The total energy of the particle is \(m a\).

Radio waves of wavelength \(\lambda\) at an angle \(\theta\) to vertical are received by a radar after reflecting from a nearby water surface and directly. If the radar records a maximum intensity, the height of antenna \(h\) from water surface can be (A) \(\frac{\lambda}{2 \cos \theta}\) (B) \(\frac{\lambda}{2 \sin \theta}\) (C) \(\frac{\lambda}{4 \sin \theta}\) (D) \(\frac{\lambda}{4 \cos \theta}\)

With propagation of longitudinal waves through a medium, the quantity transmitted is (A) Matter (B) Energy (C) Energy and matter (D) Energy, matter, and momentum

A tuning fork of known frequency \(256 \mathrm{~Hz}\) makes 5 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was (A) \(256+2 \mathrm{~Hz}\) (B) \(256-2 \mathrm{~Hz}\) (C) \(256-5 \mathrm{~Hz}\) (D) \(256+5 \mathrm{~Hz}\)
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Famous Physicists

Read Explanation

Engineering Physics

Read Explanation

Electromagnetism

Read Explanation

Electrostatics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.