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

Answer:

$F_0 = 393 N$

Explanation:

As we know that amplitude of forced oscillation is given as

$A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}$

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

$\omega_0 = \sqrt{\frac{k}{m}}$

here mass of the object is given as

$m = \frac{W}{g}$

$\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}$

$\omega_0 = 8.48 rad/s$

angular frequency of applied force is given as

$\omega = 2\pi f$

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

now we have

$0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}$

$F_0 = 393 N$