Question

A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a harmonic driving force of frequency 10.5 Hz, resulting in a forced-motion amplitude of 3.00 cm. Determine the maximum value of the driving force..

Answer 1

`F_0 = 393 N`

Step-by-step explanation:

As we know that amplitude of forced oscillation is given as

`A = (F_0)/( m(\omega^2 - \omega_0^2))`

here we know that natural frequency of the oscillation is given as

`\omega_0 = \sqrt{(k)/(m)}`

here mass of the object is given as

`m = (W)/(g)`

`\omega_0 = \sqrt{(220)/((30)/(9.81))}`

`\omega_0 = 8.48 rad/s`

angular frequency of applied force is given as

`\omega = 2\pi f`

`\omega = 2\pi(10.5) = 65.97 rad/s`

now we have

`0.03 = (F_0)/(3.06(65.97^2 - 8.48^2))`

`F_0 = 393 N`