/*! 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} Q4Q Which of the following relations... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Q4Q (page 435)

Which of the following relationships between the acceleration a and the displacement x of a particle involve SHM: (a) a=0.5x, (b) a=400x2, (c) a=20x, (d)a=-3x2?

Short Answer

Expert verified

The relationship between acceleration and displacement is a=-20x .

Step by step solution

01

The given data 

The relationships between acceleration and position are given as:

  1. a=0.5x
  2. a=400x2
  3. a=-20x
  4. a=-3x2
02

Understanding the concept of Hooke’s law

We can use Hook鈥檚 law to determine the relation between acceleration and displacement of the particle. Comparing it with the given options, we can find the correct option showing this relationship.

Formulae:

The force on a spring due to Hooke鈥檚 law F=-kx, (i)

The force due to Newton鈥檚 second law of motion, F=ma (ii)

03

Calculation of the correct acceleration and displacement relation

ma=-kxa=-kmx

Comparing this result with the options given in the problem,

The option (c) matches with the acceleration we wrote from Hook鈥檚 law.

Therefore, the relation a=-20x involves the SHM.

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 simple harmonic oscillator consists of a 0.80kgblock attached to a spring (k=200N/m). The block slides on a horizontal frictionless surface about the equilibrium pointx=0with a total mechanical energy ofrole="math" localid="1657274001354" 4.0J. (a) What is the amplitude of the oscillation? (b) How many oscillations does the block complete inrole="math" localid="1657273942909" 10s? (c) What is the maximum kinetic energy attained by the block? (d) What is the speed of the block atx=0.15m?

50.0 g stone is attached to the bottom of a vertical spring and set vibrating. The maximum speed of the stone is 15.0 cm / s and the period is 0.500 s.

(a) Find the spring constant of the spring.

(b) Find the amplitude of the motion.

(c) Find the frequency of oscillation.

The scale of a spring balance that reads from 0to 15.0 kgis12.0cm long. A package suspended from the balance is found to oscillate vertically with a frequency of 2.00 Hz.

  1. What is the spring constant?
  2. How much does the package weigh?

Find the mechanical energy of a block鈥搒pring system having a spring constant 1.3 N/ cmofand oscillation amplitude of 2.4cm.

A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 15.00 cm and mass 1.000 kg that is attached to a long thin rod of negligible mass. The pendulum swings freely about an axis perpendicular to the rod and through the end of the rod opposite the disk, as shown in Fig. 15-5 If the pendulum is to have a period of 2.000 s for small oscillations at a place where g=9.800m/s2,what must be the rod length L to the nearest tenth of a millimeter?
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Magnetism

Read Explanation

Oscillations

Read Explanation

Quantum Physics

Read Explanation

Modern Physics

Read Explanation

Fields in 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.