Question

An object of mass m (in grams) attached to a coiled spring with damping factor b (in grams per second) is pulled down a distance a (in centimeters) from its rest position and then released. Assume that the positive direction of the motion is up and the period is T (in seconds) under simple harmonic motion.

(a) Develop a model that relates the distance d of the object from its rest position after t seconds.

(b) Graph the equation found in part (a) for 5 oscillations using a graphing utility.

Given values$m=30,a=18,b=0.6,T=4$

Short answer

a) The equation can be given as $d\left(t\right)=18e^{-0.01}\cos (1.5707t)$

b) the graph can be given as

Step-by-step solution

Given information

We are given$m=30,a=18,b=0.6,T=4$

Part (a) Step 1: Find angular velocity

We get,

$$\begin{gathered} \omega =\frac{2\pi }{T} \\ \omega =\frac{2\pi }{4} \\ \omega =\frac{\pi }{2} \end{gathered}$$

Part (a) Step 2: Find the equation

We get,

$$\begin{gathered} d\left(t\right)=a{e}^{\frac{-bt}{2m}}\cos \left(\sqrt{{\omega }^2-\frac{b^2}{4m^2}}t\right) \\ d\left(t\right)=18e^{\frac{-0.6t}{2\left(30\right)}}\cos \left(\sqrt{{\left(\frac{\pi }{2}\right)}^2-\frac{0.6^2}{4{\left(30\right)}^2}}t\right) \\ d\left(t\right)=1e^{-0.01t}\cos \left(\sqrt{\left(\frac{{\pi }^2}{4}\right)-\frac{0.360}{3600}}t\right) \end{gathered}$$

Part (b) Step 1: Plot the graph

We get,

Conclusion

a) The equation can be given as$d\left(t\right)=18e^{-0.01}\cos (1.5707t)$

b) The graph can be given as