At time [latex]t=0[/latex], the displacement is the maximum of 10 cm, which calls for the cosine function. The cosine function will apply to both models.

We are given the frequency [latex]f=\frac{\omega }{2\pi }[/latex] of 0.5 cycles per second. Thus,

[latex]\begin{align}\frac{\omega }{2\pi }&=0.5 \\ \omega &=\left(0.5\right)2\pi \\ &=\pi \end{align}[/latex]

The first spring system has a damping factor of [latex]c=0.5[/latex]. Following the general model for damped harmonic motion, we have

[latex]f\left(t\right)=10{e}^{-0.5t}\cos \left(\pi t\right)[/latex]

The second spring system has a damping factor of [latex]c=0.1[/latex] and can be modeled as

[latex]f\left(t\right)=10{e}^{-0.1t}\cos \left(\pi t\right)[/latex]

Analysis of the Solution

Notice the differing effects of the damping constant. The local maximum and minimum values of the function with the damping factor [latex]c=0.5[/latex] decreases much more rapidly than that of the function with [latex]c=0.1[/latex].

Substitute the given values into the model. Recall that period is [latex]\frac{2\pi }{\omega }[/latex] and frequency is [latex]\frac{\omega }{2\pi }[/latex].

1. [latex]y=20{e}^{-0.05t}\cos \left(\frac{\pi }{2}t\right)[/latex].
2. [latex]y=2{e}^{-1.5t}\cos \left(6\pi t\right)[/latex].

Calculate the value of [latex]\omega[/latex] and substitute the known values into the model.

1. As period is [latex]\frac{2\pi }{\omega }[/latex], we have
   [latex]\begin{gathered}\frac{\pi }{6}=\frac{2\pi }{\omega } \\ \omega \pi =6\left(2\pi \right) \\ \omega =12 \end{gathered}[/latex]

   The damping factor is given as 10 and the amplitude is 7. Thus, the model is [latex]y=7{e}^{-10t}\sin \left(12t\right)[/latex].

2. As frequency is [latex]\frac{\omega }{2\pi }[/latex], we have
   [latex]\begin{gathered}20=\frac{\omega }{2\pi } \\ 40\pi =\omega \end{gathered}[/latex]

   The damping factor is given as [latex]0.2[/latex] and the amplitude is [latex]0.3[/latex]. The model is [latex]y=0.3{e}^{-0.2t}\sin \left(40\pi t\right)[/latex].

Analysis of the Solution

A comparison of the last two examples illustrates how we choose between the sine or cosine functions to model sinusoidal criteria. We see that the cosine function is at the maximum displacement when [latex]t=0[/latex], and the sine function is at the equilibrium point when [latex]t=0[/latex]. For example, consider the equation [latex]y=20{e}^{-0.05t}\cos \left(\frac{\pi }{2}t\right)[/latex] from Example 9. We can see from the graph that when [latex]t=0,y=20[/latex], which is the initial amplitude. Check this by setting [latex]t=0[/latex] in the cosine equation:

[latex]\begin{align}y&=20{e}^{-0.05\left(0\right)}\cos \left(\frac{\pi }{2}\cdot0\right) \\ &=20\left(1\right)\left(1\right) \\ &=20\end{align}[/latex]

Using the sine function yields

[latex]\begin{align}y&=20{e}^{-0.05\left(0\right)}\sin \left(\frac{\pi }{2}\cdot0\right) \\ &=20\left(1\right)\left(0\right) \\ &=0 \end{align}[/latex]

Thus, cosine is the correct function.