Log out Go to App
Go to App

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 (ω)

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

Calculate the maximum acceleration (amax)

The maximum acceleration of the particle in S.H.M. can be found using the formula: amax = ω²A Here, A is the amplitude of motion, given as 0.01 m. Plugging in the values, we have: amax = (120π)²(0.01) amax = 14400π²
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)(14400π²) Fmax = 14400π² N From the given options, the closest value is (A) 144π² N. However, the correct answer is actually 14400π² 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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 function \(\sin ^{2}(\omega t)\) represents (A) A SHM with periodic time \(\pi / \omega\) (B) A SHM with a periodic time \(2 \pi / \omega\) (C) A periodic motion with periodic time \(\pi / \omega\) (D) A periodic motion with period \(2 \pi / \omega\)

The periodic time of two oscillators are \(\mathrm{T}\) and $(5 \mathrm{~T} / 4)$ respectively. Both oscillators starts their oscillation simultaneously from the midpoint of their path of motion. When the oscillator having periodic time \(\mathrm{T}\) completes one oscillation, the phase difference between the two oscillators will be \(\ldots \ldots \ldots\) (A) \(90^{\circ}\) (B) \(112^{\circ}\) (C) \(72^{\circ}\) (D) \(45^{\circ}\)

Two sound waves are represented by $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})\( and \)\mathrm{y}=\mathrm{a} \cos (\omega \mathrm{t}-\mathrm{kx})$. The phase difference between the waves in water is \(\ldots \ldots \ldots\) (A) \((\pi / 2)\) (B) \((\pi / 4)\) (C) \(\pi\) (D) \((3 \pi / 4)\)

The equation for displacement of a particle at time \(t\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). The motion of the particle is \(\ldots \ldots\) (A) Damped motion (B) Periodic motion (C) Rotational motion (D) S.H.M.

A wave travelling along a string is described by $\mathrm{y}=0.005 \sin (40 \mathrm{x}-2 \mathrm{t})$ in SI units. The wavelength and frequency of the wave are \(\ldots \ldots \ldots\) (A) \((\pi / 5) \mathrm{m} ; 0.12 \mathrm{~Hz}\) (B) \((\pi / 10) \mathrm{m} ; 0.24 \mathrm{~Hz}\) (C) \((\pi / 40) \mathrm{m} ; 0.48 \mathrm{~Hz}\) (D) \((\pi / 20) \mathrm{m} ; 0.32 \mathrm{~Hz}\)
See all solutions

Recommended explanations on English Textbooks

Syntax

Read Explanation

Language and Social Groups

Read Explanation

Language Analysis

Read Explanation

5 Paragraph Essay

Read Explanation

Pragmatics

Read Explanation

Essay Plan

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.

Sign-up for free