Finding the Amplitude of a Trigonometric Motion Equation

[Fascheue]

Homework Statement

Find the amplitude of the motion given by x(t) = C cos ωt + D sin ωt.

Homework Equations

[/B]
x(t) = C cos ωt + D sin ωt

The Attempt at a Solution

[/B]
x(t) = C cos ωt + D sin ωt

x’(t) = -Cω sin ωt + Dω cos ωt

0 = -Cω sin ωt + Dω cos ωt

0 = -C sin ωt + D cos ωt

C sin ωt = D cos ωt

tan ωt = D/C

ωt = arctan D/C

t = (arctan D/C)/ω

x((arctan D/C)/ω) = C cos ω((arctan D/C)/ω)+ D sin ω((arctan D/C)/ω)

x((arctan D/C)/ω) = C/sqrt(1 + (D/C)^2) + D^2/Csqrt(1 + (D/C)^2)

 

[haruspex]

Homework Statement

Find the amplitude of the motion given by x(t) = C cos ωt + D sin ωt.

Homework Equations

[/B]
x(t) = C cos ωt + D sin ωt

The Attempt at a Solution

[/B]
x(t) = C cos ωt + D sin ωt

x’(t) = -Cω sin ωt + Dω cos ωt

0 = -Cω sin ωt + Dω cos ωt

0 = -C sin ωt + D cos ωt

C sin ωt = D cos ωt

tan ωt = D/C

ωt = arctan D/C

t = (arctan D/C)/ω

x((arctan D/C)/ω) = C cos ω((arctan D/C)/ω)+ D sin ω((arctan D/C)/ω)

x((arctan D/C)/ω) = C/sqrt(1 + (D/C)^2) + D^2/Csqrt(1 + (D/C)^2)

Yes, but you can simplify that a lot. (You should expect the answer to be symmetric in C and D.)

 

[Fascheue]

Yes, but you can simplify that a lot. (You should expect the answer to be symmetric in C and D.)

I’m not quite sure what you mean by symmetric in C and D, but I tried to simplify the expression and got (1 + C)(sqrt(1 + (D/C)^2))/(1 + D).

 

[haruspex]

I’m not quite sure what you mean by symmetric in C and D, but I tried to simplify the expression and got (1 + C)(sqrt(1 + (D/C)^2))/(1 + D).

That cannot be right because it adds dimensionless terms (1) to dimensioned terms (C, D). Try that again.

 

[Fascheue]

That cannot be right because it adds dimensionless terms (1) to dimensioned terms (C, D). Try that again.

Is Csqrt(1 + (D/C)^2) correct?

 

[haruspex]

Is Csqrt(1 + (D/C)^2) correct?

Yes, or in symmetric form √(C²+D²).

 

[haruspex]

Yes, or in symmetric form √(C²+D²).

Did you understand why it should be symmetric? You can swap the cos and sin around in the original equation by a combination of time reversal and phase shift. Those changes cannot alter the amplitude.

 

[Ray Vickson]

Homework Statement

Find the amplitude of the motion given by x(t) = C cos ωt + D sin ωt.

Homework Equations

[/B]
x(t) = C cos ωt + D sin ωt

The Attempt at a Solution

[/B]
x(t) = C cos ωt + D sin ωt

x’(t) = -Cω sin ωt + Dω cos ωt

0 = -Cω sin ωt + Dω cos ωt

0 = -C sin ωt + D cos ωt

C sin ωt = D cos ωt

tan ωt = D/C

ωt = arctan D/C

t = (arctan D/C)/ω

x((arctan D/C)/ω) = C cos ω((arctan D/C)/ω)+ D sin ω((arctan D/C)/ω)

x((arctan D/C)/ω) = C/sqrt(1 + (D/C)^2) + D^2/Csqrt(1 + (D/C)^2)

There is an easier way that you should know about; it is all the time in physics and engineering.

You can write a linear combination such as $S = A \sin(\theta) + B \cos(\theta)$ as a single sine or cosine, with an amplitude and a phase shift. Basically, use the addition formulas
$$\begin{array}{cl}(1):& \sin(\theta + p) = \sin(\theta) \cos(p) + \cos(\theta) \sin(p)\\
\text{or} &\\
(2):& \cos(\theta - p) = \cos(\theta) \cos(p) + \sin(\theta) \sin(p)
\end{array}$$
to re-write $S$. For example, if we use formula (1) we have $$A \sin(\theta) + B \cos(\theta) = C\cos(p) \sin(\theta) + C\sin(p) \cos(\theta),$$ so
$$ A = C \cos(p), \; B = C \sin(p) \; \Rightarrow \; C^2 = A^2 + B^2$$
If we choose the positive root $C = \sqrt{A^2+B^2},$ then we have
$$\cos(p) = \frac{A}{\sqrt{A^2+B^2}}, \; \; \sin(p) = \frac{B}{\sqrt{A^2+B^2}} $$
The signs of $A$ and $B$ will tell us which quadrant the point $(\cos(p), \sin(p))$ lies in, so we can tell which value of $p = \arcsin(B/C)$ or $p = \arctan(B/A)$ to pick.

 

[Fascheue]

Did you understand why it should be symmetric? You can swap the cos and sin around in the original equation by a combination of time reversal and phase shift. Those changes cannot alter the amplitude.

Yes, I see that now. Thanks for the help.