Asked by Ro
A weight of 30.0 N is suspended from a spring that has a force constant of 180 N/m. The system is undamped and is subjected to a harmonic driving force of frequency 11.0 Hz, resulting in a forced-motion amplitude of 5.00 cm. Determine the maximum value of the driving force.
PLEASE I NEED HELP
PLEASE I NEED HELP
Answers
Answered by
drwls
Get the natural frequency of the system
wo = sqrt (k/m)
m = 30N/g = 3.06 kg
wo= 7.67 rad/s
The forced vibration frequency is
w = 2*pi*f = 17.28 rad/s
For oscillations of this type, without damping, the amplitude A is related to the force amplitude F by
A = F/[k(1 - (w/wo)^2)]
= -0.245 F/m
You should do some reading on the subject of resonance to confirm the above formula.
Substitute 0.05m for A and solve for F
wo = sqrt (k/m)
m = 30N/g = 3.06 kg
wo= 7.67 rad/s
The forced vibration frequency is
w = 2*pi*f = 17.28 rad/s
For oscillations of this type, without damping, the amplitude A is related to the force amplitude F by
A = F/[k(1 - (w/wo)^2)]
= -0.245 F/m
You should do some reading on the subject of resonance to confirm the above formula.
Substitute 0.05m for A and solve for F
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