Question

A device consists of an object with a weight of 35.0 N hanging vertically from a spring with a spring constant of 240 N/m. There is negligible damping of the oscillating system. Applied to the system is a harmonic driving force of 13.0 Hz, which causes the object to oscillate with an amplitude of 4.00 cm. What is the maximum value of the driving force (in N)? (Enter the magnitude.)

Answer 1

The maximum value of the driving force is 942.68 N.

Explanation:

Given:

Device consists of an object with a weight =35 N

A spring with a spring constant k = 240 N/m

The system is a harmonic driving force, f = 13 Hz

The object to oscillate with an amplitude, $X_{0}=4 c m=0.04 m$

We know,  $g=9.8 \mathrm{m} / \mathrm{s}^{2}$

                 $m=\frac{35}{g}=\frac{35}{9.8}=3.57 \mathrm{kg}$

Now, given f = 13 Hz, so,

                 $\omega=2 \pi f=2 \times 3.14 \times 13=81.64 \mathrm{rad} / \mathrm{s}$

Amplitude, $X_{0}=\frac{\left(\frac{F_{0}}{k}\right)}{\sqrt{\left(1-q^{2}\right)^{2}+(2 f q)^{2}}}$ and $\varepsilon=0(\text { according to this question })$

Where, $q=\frac{\omega}{\omega_{n}}$

                $\omega_{n}=\sqrt{\frac{k}{m}}=\sqrt{\frac{240 \times 9.8}{35}}=\sqrt{\frac{2352}{35}}=\sqrt{67.2}=8.197 \mathrm{rad} / \mathrm{s}$

So, substituting the values, we get

               $q=\frac{81.64}{8.197}=9.9597$

So, for q greater than 1,

              $F_{0}=X_{0}\left(q^{2}-1\right) k=0.04 \times\left(9.9597^{2}-1\right) \times 240=0.04 \times 98.196 \times 240=942.68 N$