Chapter 5, Problem 131CR

To determine

Towrite:The model in the form of $y=\sqrt{a^2+b^2}\sin \left(Bt+C\right)$ and verify the result graphically.

Answer to Problem 131CR

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

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)

To find the amplitude of the equation (1), change the equation to the form of $y=\sqrt{a^2+b^2}\sin \left(Bt+C\right)$ .

Multiply and divide the right hand side of the equation (1) by the square root of the sum of the squares of the amplitude of the trigonometric part that is, $\sqrt{{\left(\frac{3}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}$ and simplify.

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

Now, graph both the function same window.

  

Both the graph coincide.

So, the identity is verified.