Question

The distance d (in meters) of the bob of a pendulum of mass m (in kilograms) from its rest position at time t (in seconds) is given

as$d\left(t\right)=-15e^{-0.6t/40}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}}t\right)$

(a) Describe the motion of the object.

(b) What is the initial displacement of the bob? That is, what is the displacement at t = 0?

(c) Graph the motion using a graphing utility.

(d) What is the displacement of the bob at the start of the second oscillation?

(e) What happens to the displacement of the bob as time increases without bound?

Short answer

• It is a damped motion.
• The initial displacement is 15 centimeters
• The graph of the motion is

• The displacement at the beginning of the second oscillation is 13.917.
• If the time keeps on increasing, the displacement will approach to 0 and the motion will stop.

Step-by-step solution

Part (a) Step 1. Given Information

The given equation is$d\left(t\right)=-15e^{-0.6t/40}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}}t\right)$.

Part (a) Step 2. Compare the Equation with Damped Motion

• The equation of a damped motion is$d\left(t\right)=a{e}^{-bt/2m}\cos \left(\sqrt{{\omega }^2-\frac{b^2}{4m^2}}t\right)$.
• The given equation is similar to this equation with $a=-15,b=0.6,m=20,\omega =\frac{2\mathrm{\pi }}{5}$.
• So, the motion is a damped motion.

Part (b) Step 1. Given Information

The given equation is $d\left(t\right)=-15e^{-0.6t/40}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}}t\right)$.

Part (b) Step 2. Find the Initial Displacement

• The initial displacement of the blob is $\left|a\right|$.
• Here $a=-15$.
• SO, the displacement is 15 centimeters.

Part (c) Step 1. Given Information

The given equation is$d\left(t\right)=-15e^{-0.6t/40}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}}t\right)$.

Part (c) Step 2. Plot The Function

Plot the graph of the given equation.

Part (d) Step 1. Given Information

The given equation is$d\left(t\right)=-15e^{-0.6t/40}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}}t\right)$.

Part (d) Step 2. Read From the Graph

• From the created graph, it can be seen that the second oscillation starts at the point $(4.991,-13.917)$.
• So, the displacement at the beginning of the second oscillation is 13.917.

Part (e) Step 1. Given Information

The given equation is$d\left(t\right)=-15e^{-0.6t/40}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}}t\right)$.

Part (e) Step 2. Observe From the Graph

• From the graph, it is noticed that the displacement reduces with the increasing time.
• So, if the time keeps on increasing, the displacement will approach to 0 and the motion will stop.