Question:

A 0.400-kg bob is attached to a spring of its force constant \(k=200\) N/m and this motion is subject to a resistive force, \(-bv\), which is proportional to the velocity. If the damped frequency is 99.5 % of the undamped frequency, find the value of \(b\).

Answer:

The equation of motion can be described as:

\[  \sum F = -kx – b\frac{dx}{dt} = m\frac{d^2x}{dt^2}  \]

Let us put \(\frac{dx}{dt}=x’\equiv Dx\) and \(\frac{d^2x}{dt^2}=x”\equiv D^2x\) into the above equation. The differential equation becomes

\[  x”+\frac{b}{m}x’+\frac{k}{m}x = 0  \quad \mathrm{or} \]

\[  \left(D^2+\frac{b}{m}D+\frac{k}{m}\right)x = 0 \]

Let us also put \(\omega_0 = \sqrt{\frac{k}{m}}\), which is called the rational frequency. This system is a damped oscillation, so the discriminant must be negative. Namely,

\[  \left(\frac{b}{m}\right)^2 – 4\omega^2 < 0 \ \rightarrow \ \left(\frac{b}{m}\right)^2 < 4\omega^2  \]

The solution should be

\[  D = -\frac{b}{2m} \pm \sqrt{\omega^2_0 – \frac{1}{4}\frac{b^2}{m^2}}i  \]

Therefore,

\[ x = Ae^{-\frac{m}{2m}t}\cos(\omega t + \phi)   \]

where \(\omega=\sqrt{\omega^2_0 – \frac{1}{4}\frac{b^2}{m^2}}\), which is called the damped frequency. This frequency is 99.5% of the undamped, so

\[  0.995\omega_0 = \sqrt{\omega^2_0 – \frac{1}{4}\frac{b^2}{m^2}}  \]

\[  0.010\omega^2_0 = \frac{1}{4}\frac{b^2}{m^2}  \]

\[ b = \sqrt{0.010 \times 4 \frac{k}{m} m^2}   \]

\[ b = \sqrt{0.040 km}  \]

Therefore,

\[  b = 1.789 \ \mathrm{kg/s}  \]