Question

Write y(t)=2sin 4 pi t + 5 cos 4 pi t) in the form y(t) A sin (wt + Ø) and identify the amplitude, angular frequency, and the phase shift of the spring motion.
Record your answers in the response box.

Answer 1

Expanding the desired form, we have

$A \sin(\omega t + \phi) = A \bigg(\sin(\omega t) \cos(\phi) + \cos(\omega t) \sin(\phi)\bigg)$

and matching it up with the given expression, we see that

$\begin{cases} A \sin(\omega t) \cos(\phi) = 2 \sin(4\pi t) \\ A \cos(\omega t) \sin(\phi) = 5 \cos(4\pi t) \end{cases}$

A natural choice for one of the symbols is $\omega = 4\pi$. Then

$\begin{cases} A \cos(\phi) = 2 \\ A \sin(\phi) = 5 \end{cases}$

Use the Pythagorean identity to eliminate $\phi$.

$(A\cos(\phi))^2 + (A\sin(\phi))^2 = A^2 \cos^2(\phi) + A^2 \sin^2(\phi) = A^2 (\cos^2(\phi) + \sin^2(\phi)) = A^2$

so that

$A^2 = 2^2 + 5^2 = 29 \implies A = \pm\sqrt{29}$

Use the definition of tangent to eliminate $A$.

$\tan(\phi) = \dfrac{\sin(\phi)}{\cos(\phi)} = \dfrac{A\sin(\phi)}{\cos(\phi)}$

so that

$\tan(\phi) = \dfrac52 \implies \phi = \tan^{-1}\left(\dfrac52\right)$

We end up with

$y(t) = 2 \sin(4\pi t) + 5 \cos(4\pi t) = \boxed{\pm\sqrt{29} \sin\left(4\pi t + \tan^{-1}\left(\dfrac52\right)\right)}$

where

• amplitude:

$|A| = \boxed{\sqrt{29}}$

• angular frequency:

$\boxed{4\pi}$

• phase shift:

$4\pi t + \tan^{-1}\left(\dfrac 52\right) = 4\pi \left(t + \boxed{\frac1{4\pi} \tan^{-1}\left(\frac52\right)}\,\right)$