   Chapter 13, Problem 28P

Chapter
Section
Textbook Problem

A harmonic oscillator is described by the function x(t) = (0.200 m) cos (0.350t). Find the oscillator’s (a) maximum velocity and (b) maximum acceleration. Find the oscillator’s (c) position, (d) velocity, and (c) acceleration when t = 2.00 s.

a)

To determine
The maximum velocity of the harmonic oscillator.

Explanation

Given info: The position of the harmonic oscillator is described by x=(0.200m)cos(0.350t) .

Explanation:

The position of the object is:

x=(0.200m)cos(0.350t)

Here,

x is the position

t is time taken

The general expression of position as a function of time for an object moving in simple harmonic motion is given by,

x=Acos(ωt)

Here,

A is the amplitude of the simple harmonic motion

ω is the angular frequency

From above equations,

The amplitude A=0.200m

The angular frequency ω=0.350rads-1

To find the expression for velocity differentiate the equation of x with respect to t.

v=Aωsin(ωt)

The maximum velocity will corresponds to minimum value of sin(ωt) , which is -1

b)

To determine
The maximum value of acceleration.

c)

To determine
The position of the oscillator.

d)

To determine
The velocity of the harmonic oscillator.

e)

To determine
The value of acceleration.

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