/*! 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 1374 A particle having mass \(1 \math... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 1374

A particle having mass \(1 \mathrm{~kg}\) is executing S.H.M. with an amplitude of \(0.01 \mathrm{~m}\) and a frequency of \(60 \mathrm{hz}\). The maximum force acting on this particle is \(\ldots \ldots . . \mathrm{N}\) (A) \(144 \pi^{2}\) (B) \(288 \pi^{2}\) (C) \(188 \pi^{2}\) (D) None of these. (A) \(x=a \sin 2 p \sqrt{(\ell / g) t}\) (B) \(x=a \cos 2 p \sqrt{(g / \ell) t}\) (C) \(\mathrm{x}=\mathrm{a} \sin \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\) (D) \(\mathrm{x}=\mathrm{a} \cos \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\)

Short Answer

Expert verified
The maximum force acting on the particle is \(14400\pi^2\) N, so the correct answer is (D) None of these.

Step by step solution

01

Calculate the angular frequency (蠅)

To find the angular frequency, we'll use the formula: 蠅 = 2蟺f Here, f is the frequency of the motion, which is given as 60 Hz. Plugging in the values, we have: 蠅 = 2蟺(60) 蠅 = 120蟺
02

Calculate the maximum acceleration (amax)

The maximum acceleration of the particle in S.H.M. can be found using the formula: amax = 蠅虏A Here, A is the amplitude of motion, given as 0.01 m. Plugging in the values, we have: amax = (120蟺)虏(0.01) amax = 14400蟺虏
03

Calculate the maximum force (Fmax)

Using Newton's second law of motion (F = ma), we can find the maximum force acting on the particle: Fmax = m*amax Here, m is the mass of the particle, given as 1 kg. Plugging in the values, we have: Fmax = (1)(14400蟺虏) Fmax = 14400蟺虏 N From the given options, the closest value is (A) 144蟺虏 N. However, the correct answer is actually 14400蟺虏 N, none of the options provided are exactly accurate. Thus, the answer is (D) None of these.

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.

Angular Frequency
In simple harmonic motion (SHM), angular frequency (\(\omega\)) plays a crucial role in describing how quickly the oscillations are happening. It is directly tied to the frequency of the oscillation, which is how many cycles or vibrations occur per unit time. The relationship between the ordinary frequency (\(f\)) and the angular frequency is given by:\[ \omega = 2\pi f \]
  • Here, \(\pi\) is a constant approximately equal to 3.14159.
  • The factor of \(2\pi\) arises because a full cycle of a circular motion (or one complete oscillation) corresponds to a traversal of \(2\pi\) radians.
For the given exercise, since the frequency \(f\) is 60 Hz, we calculate angular frequency as \(\omega = 120\pi\) rad/s. This means the particle completes 120\pi radians per second, which tells us how fast the particle repeats its oscillatory path.
Maximum Acceleration
In SHM, the maximum acceleration (\(a_{max}\)) occurs when the particle reaches its maximum displacement (amplitude). This is because acceleration is proportional to displacement in SHM but in the opposite direction, according to Hooke's law.The formula to find the maximum acceleration is given by:\[ a_{max} = \omega^2 A \]
  • Here, \(A\) is the amplitude of the motion.
  • \(\omega^2\) accentuates the quadratic influence of the angular frequency on acceleration.
Given that \(\omega = 120\pi\) rad/s and \(A = 0.01\) m, the maximum acceleration calculated was: \[a_{max} = (120\pi)^2 \times 0.01 = 14400\pi^2 \text{ m/s}^2\]This high acceleration reflects the vigorous nature of rapid oscillations.
Newton's Second Law
Newton's second law of motion is a fundamental principle that connects the concepts of force, mass, and acceleration through the equation:\[ F = ma \]
  • Here, \(F\) represents force, \(m\) is the mass of the object, and \(a\) is its acceleration.
  • It states that the force exerted by or on an object is directly proportional to the acceleration inversely proportional to its mass.
In the context of this exercise, the law helps us understand how the mass affects the force experienced by the particle during SHM. Given a mass of 1 kg and the calculated maximum acceleration, the second law allows us to find the force that acts on the particle at its extreme positions.
Force Calculation
Calculating the maximum force (\(F_{max}\)) in SHM involves using the determined values of maximum acceleration and mass within the framework provided by Newton's second law: \[ F_{max} = m \times a_{max}\]Given that the mass (\(m\)) of the particle is 1 kg and the maximum acceleration (\(a_{max}\)) is \(14400\pi^2\) m/s虏, the force acting on the particle reaches: \[ F_{max} = 1 \times 14400\pi^2 = 14400\pi^2 \text{ N}\]
  • This result underscores how force is a product of the combined effects of mass and acceleration.
  • The particle experiences this maximal force at the turnaround points of its motion where displacement and acceleration are greatest.
Therefore, while the options presented in the original problem list a force of \(144\pi^2\) N, the correctly calculated maximum force based on provided parameters is actually \(14400\pi^2\) N.

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 amplitude for a S.H.M. given by the equation \(\mathrm{x}=3 \sin 3 \mathrm{pt}+4 \cos 3 \mathrm{pt}\) is \(\ldots \ldots \ldots \ldots \mathrm{m}\) (A) 5 (B) 7 (C) 4 (D) \(3 .\)

Twenty four tuning forks are arranged in such a way that each fork produces 6 beats/s with the preceding fork. If the frequency of the last tuning fork is double than the first fork, then the frequency of the second tuning fork is \(\ldots \ldots\) (A) 132 (B) 138 (C) 276 (D) 144

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) \(2.6 \times 10^{3}\)

The equation for displacement of a particle at time \(\mathrm{t}\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). If the mass of the particle is \(5 \mathrm{gm}\), then the total energy of the particle is \(\ldots \ldots \ldots\) erg (A) 250 (B) 125 (C) 500 (D) 375

If the equation for a transverse wave is \(\mathrm{y}=\mathrm{A} \operatorname{Sin} 2 \pi\) \(\\{(1 / \mathrm{T})-(\mathrm{x} / \lambda)\\}\). then for what wavelength will the maximum velocity of the particle be double the wave velocity ? (A) \((\pi \mathrm{A} / 4)\) (B) \((\pi \mathrm{A} / 2)\) (C) \(\pi \mathrm{A}\) (D) \(2 \pi \mathrm{A}\)
See all solutions

Recommended explanations on English Textbooks

Rhetoric

Read Explanation

Synthesis Essay

Read Explanation

International English

Read Explanation

Creative Story

Read Explanation

Language and Social Groups

Read Explanation

Essay Prompts

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.