Question

An unhappy \({\bf{0}}.{\bf{300}} - {\bf{kg}}\) rodent, moving on the end of a spring with force constant \({\bf{k}} = {\bf{2}}.{\bf{50}}{\rm{ }}{\bf{N}}/{\bf{m}}\), is acted on by a damping force \({F_x} = - b{v_x}\). (a) If the constant \(b\) has the value \({\bf{0}}.{\bf{900}}{\rm{ }}{\bf{kg}}/{\bf{s}}\), what is the frequency of oscillation of the rodent? (b) For what value of the constant \(b\) will the motion be critically damped?

Short answer

(a) The frequency of oscillation of the rodent is \(0.393\,{\rm{Hz}}\)

Step-by-step solution

Identification of given data

Mass of an unhappy rodent \(m = 0.3\,{\rm{kg}}\)

Force constant of spring \(k = \,\,{\rm{2}}{\rm{.5}}\,{\rm{N/m}}\)

Constant \(b = 0.9\,{\rm{kg/s}}\)

 Step 2: Significance of frequency of damped harmonic oscillation  

The damping frequency is the frequency at which the system remains in equilibrium and at which a complete reduction in vibration amplitude is possible.

It is expressed as,

\(\omega = \sqrt {\frac{k}{m} - \frac{{{b^2}}}{{4{m^2}}}} \) …(i)

(a) Determining the frequency of oscillation of the rodent

Using equation (i)

\(\omega = \sqrt {\frac{k}{m} - \frac{{{b^2}}}{{4{m^2}}}} \)

Substitute all the values in equation (i)

\(\begin{array}{c}\omega = \sqrt {\frac{{2.5\,{\rm{N/m}}}}{{0.3\,{\rm{kg}}}} - \frac{{{{\left( {0.9\,{\rm{kg/s}}} \right)}^2}}}{{4 \times {{\left( {0.3\,{\rm{kg}}} \right)}^2}}}} \\ = 2.47\,{\rm{rad/s}}\end{array}\)

So frequency will be,

\(\begin{array}{c}f = \frac{\omega }{{2\pi }}\\ = \frac{{2.47\,{\rm{rad/s}}}}{{2\pi }}\\ = 0.393\,{\rm{Hz}}\end{array}\)

Hence, the frequency of oscillation of the rodent is \(0.393\,{\rm{Hz}}\)