Chapter 5, Problem 132CR

To determine

Tocalculate:The amplitude of the oscillations of the weight and the frequency of the oscillations of the weight.

Answer to Problem 132CR

The required amplitude is $\frac{\sqrt{10}}{2}$ .

The required frequency of oscillations is $\frac{2\pi }{{\tan }^{-1}\left(-\frac{1}{3}\right)}$ .

Explanation of Solution

Given information:

The model: $y=\frac{3}{2}\sin 8t-\frac{1}{2}\cos 8t$

Calculation:

Given the equation of displacement of the harmonic system is:

  $y=\frac{3}{2}\sin 8t-\frac{1}{2}\cos 8t$...............(1)

Multiply and divide the right hand side of the above expression by $\sqrt{3^2+{\left(-1\right)}^2}$ .

  $\begin{array}{c}y=0.5\left(\sqrt{3^2+{\left(-1\right)}^2}\right)\left[\frac{3}{\left(\sqrt{3^2+{\left(-1\right)}^2}\right)}\sin 8t-\frac{1}{\left(\sqrt{3^2+{\left(-1\right)}^2}\right)}\cos 8t\right]\\ y=\frac{1}{2}\left(\sqrt{10}\right)\left(\frac{3}{\sqrt{10}}\right)\sin 8t-\frac{1}{\sqrt{10}}\cos 8t\\ y=\frac{\sqrt{10}}{2}\left(\frac{3}{\sqrt{10}}\sin 8t-\frac{1}{\sqrt{10}}\cos 8t\right)\end{array}$

Here, the amplitude of the sine function is 1.

So, substitute 1 for $\left(\sin \left(8t-{\cos }^{-1}\left(\frac{\frac{3}{2}}{\sqrt{{\left(\frac{3}{2}\right)}^2+{\left(\frac{-1}{2}\right)}^2}}\right)\right)\right)$ in the above expression.

  $\begin{array}{c}y=\sqrt{{\frac{3}{2}}^2+{\left(-\frac{1}{2}\right)}^2}\cdot \left(1\right)\\ =\sqrt{\frac{9}{4}+\frac{1}{4}}\\ =\frac{\sqrt{10}}{2}\end{array}$

Hence, the required amplitude is $\frac{\sqrt{10}}{2}$ .

Assume that $\frac{3}{\sqrt{10}}=\cos \alpha$ and $-\frac{1}{\sqrt{10}}=\sin \alpha$ .

Substitute $\cos \alpha$ for $\frac{3}{\sqrt{10}}$ and $\sin \alpha$ for $-\frac{1}{\sqrt{10}}$ in the above expression.

  $\begin{array}{l}y=\frac{\sqrt{10}}{2}\left(\cos \alpha \sin 8t+\sin \alpha \cos 8t\right)\\ y=\frac{\sqrt{10}}{2}\left(\sin 8t\cos \alpha +\cos 8t\sin \alpha \right)\end{array}$

Use the sum and difference formula $\sin u\cos v+\cos u\sin v=\sin \left(u+v\right)$ on the above expression.

  $y=\frac{\sqrt{10}}{2}\sin \left(8t+\alpha \right)$...............(1)

Know that $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$ .

Substitute $-\frac{1}{\sqrt{10}}$ for $\sin \alpha$ and $\frac{3}{\sqrt{10}}$ for $\cos \alpha$ in the above expression.

  $\begin{array}{c}\tan \alpha =\frac{-\frac{1}{\cancel{\sqrt{10}}}}{\frac{3}{\cancel{\sqrt{10}}}}\\ \tan \alpha =-\frac{1}{3}\\ \alpha ={\tan }^{-1}\left(-\frac{1}{3}\right)\end{array}$

Now, substitute ${\tan }^{-1}\left(-\frac{1}{3}\right)$ for $\alpha$ in the expression (1).

  $y=\frac{\sqrt{10}}{2}\sin \left[8t+{\tan }^{-1}\left(-\frac{1}{3}\right)\right]$...............(2)

The equation (2) is in the form of the equation $y=\sqrt{a^2+b^2}\sin \left(Bt+C\right)$ .

In the equation (2), the time period $t$ can be written as,

  $t={\tan }^{-1}\left(-\frac{1}{3}\right)$

To find the required frequency of oscillations, divide $2\pi$ by the time period $t$ that is, $\frac{2\pi }{t}$ .

Substitute ${\tan }^{-1}\left(-\frac{1}{3}\right)$ for t in the above expression.

  $\frac{2\pi }{t}=\frac{2\pi }{{\tan }^{-1}\left(-\frac{1}{3}\right)}$

Hence, the required frequency of oscillations is $\frac{2\pi }{{\tan }^{-1}\left(-\frac{1}{3}\right)}$ .