Question

engineers can determine properties of a structure that is modeled as a damped spring oscillator, such as a bridge, by applying a driving force to it. a weakly damped spring oscillator of mass 0.236 kg is driven by a sinusoidal force at the oscillators resonance frequency of 28.8 hz.
find the value of the spring constant k.
the amplitude of the driving force is 0.505 n and the amplitude of the oscillators steady - state motion in response to this driving force is 0.997 m.
what is the oscillators damping constant b?

Explanation:

Step1: Find the spring constant \(k\)

At resonance, the angular frequency \(\omega = 2\pi f\), and for a spring - mass system \(\omega=\sqrt{\frac{k}{m}}\). So, \(k = m\omega^{2}=m(2\pi f)^{2}\).
Substitute \(m = 0.236\space kg\) and \(f = 28.8\space Hz\) into the formula:
\(k=0.236\times(2\pi\times28.8)^{2}\)
\(k = 0.236\times4\pi^{2}\times28.8^{2}\)
\(k=0.236\times4\times9.87\times829.44\)
\(k = 0.236\times32771.35\)
\(k\approx7730\space N/m\)

Step2: Find the damping constant \(b\)

At resonance, the amplitude \(X=\frac{F_{m}}{b\omega}\). We know \(F_{m} = 0.505\space N\), \(X = 0.997\space m\), and \(\omega=2\pi f=2\pi\times28.8\space rad/s\approx 180.96\space rad/s\)
From \(b=\frac{F_{m}}{X\omega}\)
Substitute the values: \(b=\frac{0.505}{0.997\times180.96}\)
\(b=\frac{0.505}{180.42}\)
\(b\approx0.0028\space kg/s\)

Answer:

\(k\approx7730\space N/m\)
\(b\approx0.0028\space kg/s\)