EXAMPLE 13.6 The Vibrating Object-Spring System Goal Identify the physical param

Question

EXAMPLE 13.6 The Vibrating Object-Spring System Goal Identify the physical parameters of a harmonic oscillator from its mathematical description.

Problem (a) Find the amplitude, frequency, and period of motion for an object vibrating at the end of a horizontal spring if the equation for its position as a function of time is (b) Find the maximum magnitude of the velocity and acceleration. (c) What are the position, velocity, and acceleration of the object after 1.00 s has elapsed?

Strategy In part (a) the amplitude and frequency can be found by comparing the given equation with the standard form, matching up the numerical values with the corresponding terms in the standard form. In part (b) the maximum speed will occur when the sine function equals 1 or ?1, the extreme values of the sine function (and similarly for the acceleration and the cosine function). In each case, find the magnitude of the expression in front of the trigonometric function. Part (c) is just a matter of substituting values into the necessary equations.




Use the worked example above to help you solve this problem. (a) Find the amplitude, frequency, and period of motion for an object vibrating at the end of a horizontal spring if the equation for its position as a function of time is the following.


If the object-spring system is described by x = (0.345 m) cos (1.00t), find the following. (a) the amplitude, the angular frequency, the frequency, and the period A = m
? = rad/s
f = Hz
T = s
(b) the maximum magnitudes of the velocity and the acceleration vmax = m/s
amax = m/s2
(c) the position, velocity, and acceleration when t = 0.250 s x = m
v = m/s
a = m/s2 EXAMPLE 13.6 The Vibrating Object-Spring System


If the object-spring system is described by x = (0.345 m) cos (1.00t), find the following. (a) the amplitude, the angular frequency, the frequency, and the period A = m
? = rad/s
f = Hz
T = s
(b) the maximum magnitudes of the velocity and the acceleration vmax = m/s
amax = m/s2
(c) the position, velocity, and acceleration when t = 0.250 s x = m
v = m/s
a = m/s2

Explanation / Answer

A=0.345 m w=1 rad/s f=w/2*pi=0.159 hz T=6.28 s v(max)=A*w=0.345 m/s a(max)=A*w^2=0.345 m/s^2 at t=0.25 s x=0.334 m v=-0.345*sin(1*t) v=-0.085 m/s a=-0.345*cos(1*t) a=-0.334 m/s^2

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