/*! 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} 9P The function x=(6.0聽m)cos[3蟺ra... [FREE SOLUTION] | 91影视

91影视

The function x=(6.0m)cos[3蟺谤补诲/st+3rad] gives the simple harmonic motion of a body. At data-custom-editor="chemistry" t=2.0s,

  1. What is the displacement of the motion?
  2. What is the velocity of the motion?
  3. What is the acceleration of the motion?
  4. What is the phase of the motion?
  5. What is the frequency of the motion?
  6. What is the period of the motion?

Short Answer

Expert verified
  1. Displacement of the motion is 3 m.
  2. Velocity of the motion is -49 m/s.
  3. Acceleration of the motion is -270m/s2.
  4. Phase of the motion is 20 rad.
  5. Frequency of the motion is 1.5 Hz.
  6. Period of the motion is 0.67 sec.

Step by step solution

01

Stating the given data

The position function is x=6cos3蟺迟+3.

02

Understanding the concept of motion

The velocity function can be found by differentiating the position equation with respect to time and acceleration, and the acceleration is a derivative of the velocity with respect to time. To find the frequency, time period, and phase, we can compare the given equation of position with the standard equation.

Formulae:

The time period of a body in motion

T=1f (i)

The velocity of a body

v=dxdt (ii)

The acceleration of a body

a=dvdt (iii)

Angular frequency of a body in oscillation

=2f (iv)

03

a) Calculation of displacement

Using the given equation of displacement of motion att=2sec, we get the displacement of the motion as

x=6cos32+3=3m

Hence, the value of displacement is 3 m.

04

b) Calculation of velocity

Using equation (ii), we get the velocity by differentiating the given displacement equation with respect to time.

v=ddt6cos3蟺迟+3=-63蟺蝉颈苍3蟺迟+3................at=2sec=-49m/s

Hence, the value of velocity is -49 m/s.

05

c) Calculation of acceleration

Using equation (ii) and from equation (a), we get the velocity by differentiating the given displacement equation with respect to time.

a=ddt-63蟺蝉颈苍3蟺迟+3=-633蟺肠辞蝉3蟺迟+3=-270m/s2

Hence, the value of acceleration is -270m/s2.

06

d) Calculation of phase

By comparing withthedisplacement equation of simple harmonic motion, the phase is calculated as
=32+3=19.987rad19.9rad20rad

Hence, the phase value is 20 rad.

07

e) Calculation of frequency

From equation of displacement,=3rad/sec.

Hence, using equation (iv), the frequency of the motion is calculated as

f=2=3/2=1.5Hz

Hence, the value of frequency is 1.5 Hz.

08

f) Calculation of period

Using equation (i), we get the period of motion as

T=11.5=0.67sec

Hence, the value of period is 0.67 sec.

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 loudspeaker diaphragm is oscillating in simple harmonic motion with a frequency of440Hzand a maximum displacement of0.75mm.

  1. What is the angular frequency?
  2. What is the maximum speed?
  3. What is the magnitude of the maximum acceleration?

Question: A thin uniform rod (mass m =0.50 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period of1.5 sand an angular amplitude of 100.

  1. What is the length of the rod?
  2. What is the maximum kinetic energy of the rod as it swings?

What is the frequency of a simple pendulum 2.0mlong (a) in a room, (b) in an elevator accelerating upward at a rate of role="math" localid="1657259780987" 2.0m/s2, and (c) in free fall?

A block weighing 10.0 Nis attached to the lower end of a vertical spring (k=200.0N/m), the other end of which is attached to a ceiling. The block oscillates vertically and has a kinetic energy of 2.00 Jas it passes through the point at which the spring is unstretched. (a) What is the period of the oscillation? (b) Use the law of conservation of energy to determine the maximum distance the block moves both above and below the point at which the spring is unstretched. (These are not necessarily the same.) (c) What is the amplitude of the oscillation? (d) What is the maximum kinetic energy of the block as it oscillates?

A damped harmonic oscillator consists of a block (m=2.00kg), a spring (k=10.0N/m), and a damping force (F=-bv). Initially, it oscillates with amplitude of25.0cm; because of the damping, the amplitude falls to three-fourths of this initial value at the completion of four oscillations. (a) What is the value of b? (b) How much energy has been 鈥渓ost鈥 during these four oscillations?

See all solutions

Recommended explanations on Physics Textbooks

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.