d'x kx = m di dx 2 = ma, → F, sin ot-b dt (15.34) Again, the solution of this equation is rather lengthy and will not be presented. After the driving force on an initially stationary object begins to act, the ampli- tude of the oscillation will increase. The system of the oscillator and the surround- ing medium is a nonisolated system: work is done by the driving force, such that the vibrational energy of the system (kinetic energy of the object, elastic potential energy in the spring) and internal energy of the object and the medium increase. After a sufficiently long period of time, when the energy input per cycle from the driving force equals the amount of mechanical energy transformed to internal energy for cach cycle, a steady-state condition is reached in which the oscillations proceed with constant amplitude. In this situation, the solution of Equation 15.34 is x = A cos (wt + 6) (15.35) F/m bo (15.36) (² – w,)²

Physics for Scientists and Engineers: Foundations and Connections
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Author:Katz, Debora M.
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Chapter16: Oscillations
Section: Chapter Questions
Problem 74PQ: The total energy of a simple harmonic oscillator with amplitude 3.00 cm is 0.500 J. a. What is the...
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Considering an undamped, forced oscillator (b = 0), show that Equation 15.35 is a solution of Equation 15.34, with an amplitude given by Equation 15.36.

d'x
kx = m
di
dx
2 = ma, → F, sin ot-b
dt
(15.34)
Again, the solution of this equation is rather lengthy and will not be presented.
After the driving force on an initially stationary object begins to act, the ampli-
tude of the oscillation will increase. The system of the oscillator and the surround-
ing medium is a nonisolated system: work is done by the driving force, such that
the vibrational energy of the system (kinetic energy of the object, elastic potential
energy in the spring) and internal energy of the object and the medium increase.
After a sufficiently long period of time, when the energy input per cycle from the
driving force equals the amount of mechanical energy transformed to internal
energy for cach cycle, a steady-state condition is reached in which the oscillations
proceed with constant amplitude. In this situation, the solution of Equation 15.34 is
x = A cos (wt + 6)
(15.35)
Transcribed Image Text:d'x kx = m di dx 2 = ma, → F, sin ot-b dt (15.34) Again, the solution of this equation is rather lengthy and will not be presented. After the driving force on an initially stationary object begins to act, the ampli- tude of the oscillation will increase. The system of the oscillator and the surround- ing medium is a nonisolated system: work is done by the driving force, such that the vibrational energy of the system (kinetic energy of the object, elastic potential energy in the spring) and internal energy of the object and the medium increase. After a sufficiently long period of time, when the energy input per cycle from the driving force equals the amount of mechanical energy transformed to internal energy for cach cycle, a steady-state condition is reached in which the oscillations proceed with constant amplitude. In this situation, the solution of Equation 15.34 is x = A cos (wt + 6) (15.35)
F/m
bo
(15.36)
(² – w,)²
Transcribed Image Text:F/m bo (15.36) (² – w,)²
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