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Dmitrij [34]
3 years ago
7

Damping is negligible for a 0.135-kg object hanging from a light, 6.30-N/m spring. A sinusoidal force with an amplitude of 1.70

N drives the system. At what frequency will the force make the object vibrate with an amplitude of 0.480 m?
Physics
1 answer:
IceJOKER [234]3 years ago
8 0

Answer:

0.72 Hz minimum frequency

Explanation:

When the damping is negligible,Amplitude is given as

A = (F/m)/[\sqrt{(\omega ^{^{2}}-\omega _{o}^{2}})^2

here \omega _{o}^{2}= k/m = (6.30)/(0.135) = 46.67 N/m kg

F / mA = 1.70/(0.135)(0.480) = 26.2 N/m kg

From the above equation , rearranging for ω,

\omega ^{2}= \omega _{o}^{2}\pm F/m

⇒ ω² =46.67 ± 26.2 = 72.87 or 20.47

⇒ ω = 8.53 or 4.52 rad/s

Frequency = f

ω=2 π f

⇒ f = ω / 2π =  8.53 /6.28  or 4.52 / 6.28 = 1.36 Hz or 0.72 Hz

The lower frequency is 0.72 Hz and higher is 1.36 Hz

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