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Physics For Scientists & Engineers
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Short Answer

A 2.00kg object attached to a spring moves without friction (b=0) and is driven by an external force given by the expression F=3.00(sin2πt), where F is in Newton’s and t is in seconds. The force constant of the spring is 20.0Nm-1. Find

(a) The resonance angular frequency of the system,

(b) The angular frequency of the driven system, and

(c) The amplitude of the motion.

(c) The amplitude of the motion is A = 5.09 cm.

See the step by step solution

Step by Step Solution

Step 1: Identification of the given data

The given data can be listed below as,

  • The external force equation is F=3.00sin2πt.
  • The mass of object is m=2.00kg.

Step 2: Significance of the amplitude

Amplitude of driven oscillator with no damping is given by:

A=F0m(ω2-ω02)

Step 3: Determination of the amplitude of the motion.

Part (c):

We have to find the resonance angular frequency of the system:

By using concept and formula from step (1), we get

A=F0mω2-ω02

Referring to subparts (a) and (b) of the SID: 947271-15-15.7-947271-15-53P-a and 947271-15-15.7-947271-15-53P-b

Substitute all the value in the above equation,

A=3Nm-12kg6.28s-12-3.16s-12A=0.0509mA=5.09cm

Hence the amplitude of the motion is A = 5.09 cm.

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