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45 votes
45 votes
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.

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User Ronze
by
3.0k points

1 Answer

12 votes
12 votes

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√(29)

Use the definition of tangent to eliminate
A.


\tan(\phi) = (\sin(\phi))/(\cos(\phi)) = (A\sin(\phi))/(\cos(\phi))

so that


\tan(\phi) = \frac52 \implies \phi = \tan^(-1)\left(\frac52\right)

We end up with


y(t) = 2 \sin(4\pi t) + 5 \cos(4\pi t) = \boxed{\pm√(29) \sin\left(4\pi t + \tan^(-1)\left(\frac52\right)\right)}

where

amplitude:


|A| = \boxed{√(29)}

angular frequency:


\boxed{4\pi}

phase shift:


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

User Kelvin Omereshone
by
2.9k points
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