/*! 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 20 A particle is executing SHM with... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 20

A particle is executing SHM with an amplitude of \(4 \mathrm{~cm}\). At the mean position, velocity of the particle is \(10 \mathrm{~cm} / \mathrm{s}\). The distance of the particle from the mean position when its speed becomes \(5 \mathrm{~cm} / \mathrm{s}\) is (A) \(\sqrt{3} \mathrm{~cm}\) (B) \(\sqrt{5} \mathrm{~cm}\) (C) \(2 \sqrt{3} \mathrm{~cm}\) (D) \(2 \sqrt{5} \mathrm{~cm}\)

Short Answer

Expert verified
The distance of the particle from the mean position when its speed becomes 5 cm/s is (A) \(\sqrt{3} \mathrm{~cm}\).

Step by step solution

01

Understanding the velocity formula in SHM

We know that the velocity in Simple Harmonic Motion (SHM) can be expressed as: \(v = \omega \sqrt{A^2 - x^2}\), where v is the velocity, x is the distance of the particle from its mean position, A is the amplitude of the motion and 蠅 is angular frequency of the SHM.
02

Calculating the angular frequency

We are given the velocity at the mean position (v = 10 cm/s) and the amplitude (A = 4 cm). When the particle is at its mean position, x = 0. Plugging in the given values, \(v = \omega \sqrt{A^2 - x^2}\) becomes \(10 = \omega \sqrt{4^2 - 0^2}\). Solving for omega, \[\omega = \frac{10}{\sqrt{4^2}} = \frac{10}{4} = \frac{5}{2} \text{ s}^{-1}\].
03

Using the given speed to find the distance from mean position

Now, we are given that the speed of the particle is 5 cm/s. Using the same formula as earlier, but with the new velocity (v = 5 cm/s) and the calculated angular frequency (蠅 = 5/2 s^{-1}), we can solve for x. \(5 = \frac{5}{2} \sqrt{4^2 - x^2}\)
04

Simplifying the equation

Dividing by 5 and squaring both sides, we get: \[\frac{1}{4} = 1 - \frac{x^2}{4^2}\]
05

Solving for x

Next, we solve for x: \[x^2 = 4^2 (1 - \frac{1}{4})\] \[x^2 = 4^2 (\frac{3}{4})\] \[x^2 = 3^2\] So, x = 3 cm. Now, let's check which one of the options matches our answer. Based on our solution, the answer is (A) \(\sqrt{3} \mathrm{~cm}\).

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!

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

A tuning fork of frequency \(340 \mathrm{~Hz}\) is vibrated just above a cylindrical tube of length \(120 \mathrm{~cm}\). Water is slowly poured in the tube. If the speed of sound is 340 \(\mathrm{m} / \mathrm{s}\), then the minimum height of water required for resonance is (A) \(25 \mathrm{~cm}\) (B) \(45 \mathrm{~cm}\) (C) \(75 \mathrm{~cm}\) (D) \(95 \mathrm{~cm}\)

In a simple harmonic oscillator, at the mean position (A) kinetic energy is minimum, potential energy is maximum. (B) both kinetic and potential energies are maximum. (C) kinetic energy is maximum, potential energy is minimum. (D) both kinetic and potential energies are minimum.

A metal wire of linear mass density of \(9.8 \mathrm{~g} / \mathrm{m}\) is stretched with a tension of \(10 \mathrm{~kg}-\mathrm{wt}\) between two rigid supports 1 meter apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency \(n .\) The frequency \(\mathrm{n}\) of the alternating source is (A) \(50 \mathrm{~Hz}\) (B) \(100 \mathrm{~Hz}\) (C) \(200 \mathrm{~Hz}\) (D) \(25 \mathrm{~Hz}\)

A pipe open at both ends has a fundamental frequency \(f\) in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now (A) \(\frac{3 f}{4}\) (B) \(2 f\) (C) \(f\) (D) \(\frac{f}{2}\)

The general wave equation can be written as \(y=m(x-v t), x \in\left[v t, v t+\frac{a}{2}\right]$$y=-m[(x-v t)-a], x \in\left[v t+\frac{a}{2}, v t+a\right]\)
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Famous Physicists

Read Explanation

Thermodynamics

Read Explanation

Quantum Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Waves Physics

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.