Question

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?

Answer 1

0.72 Hz minimum frequency

Step-by-step 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