# Finding the amplitude

• Nov 13th 2010, 10:42 AM
Finding the amplitude
I have

$\displaystyle y=5 sin (4t) + 9 Cos (4t)$
i am told to put it in the form; $\displaystyle R Sin (4t + \alpha)$where $\displaystyle 0 < \alpha < (\pi/2)$

and the period is $\displaystyle t= 2 \pi/w$ = 1.57 secs

how to you find the amplitude???
• Nov 13th 2010, 10:50 AM
CaptainBlack
Quote:

I have

$\displaystyle y=5 sin (4t) + 9 Cos (4t)$
i am told to put it in the form; $\displaystyle R Sin (4t + \alpha)$where $\displaystyle 0 < \alpha < (\pi/2)$

and the period is $\displaystyle t= 2 \pi/w$ = 1.57 secs

how to you find the amplitude???

Trigonometry!

Put $\displaystyle \alpha=\arctan(9/5)$ then:

$\displaystyle y=\sqrt{5^2+9^2} \cos(\alpha) \sin (4t) + \sqrt{5^2+9^2} \sin(\alpha) \cos (4t)$

etc..

CB
• Nov 13th 2010, 11:00 AM
Quote:

Originally Posted by CaptainBlack
Trigonometry!

Put $\displaystyle \alpha=\arctan(9/5)$ then:

$\displaystyle y=\sqrt{5^2+9^2} \cos(\alpha) \sin (4t) + \sqrt{5^2+9^2} \sin(\alpha) \cos (4t)$

etc..

CB

Therefore; $\displaystyle \alpha=\arctan(9/5)= \arctan(1.8)= 60.94$ in degrees, then:

Convert to radians; $\displaystyle 60.94 (\pi/180)=1.06$

Now;

$\displaystyle y=\sqrt{5^2+9^2} \cos(\alpha) \sin (4t) + \sqrt{5^2+9^2} \sin(\alpha) \cos (4t)$

$\displaystyle y=\sqrt{106} \cos(1.06) \sin (4(1.57)) + \sqrt{106} \sin(1.06) \cos (4(1.57))$

$\displaystyle y=10.29 .(0.489) \sin (6.28) + 10.29 .(0.873) \cos (6.28)$

$\displaystyle y=10.29 . (0.489) . (-0.003) + 10.29 . (0.873) . (0.99)$

$\displaystyle y=-0.0151 + 8.893$

$\displaystyle y=8.88$.....

is this correct?, would y be the answer for the amplitude?
• Nov 13th 2010, 12:10 PM
Found this;

where y from the question = Ao on the diagram above

x-axis would be time (in secs)
• Nov 13th 2010, 12:31 PM
CaptainBlack
Quote:

Therefore; $\displaystyle \alpha=\arctan(9/5)= \arctan(1.8)= 60.94$ in degrees, then:

Convert to radians; $\displaystyle 60.94 (\pi/180)=1.06$

Now;

$\displaystyle y=\sqrt{5^2+9^2} \cos(\alpha) \sin (4t) + \sqrt{5^2+9^2} \sin(\alpha) \cos (4t)$

$\displaystyle y=\sqrt{5^2+9^2} \sin (4t+\alpha)$

I don't know what you think the rest means:-

Quote:

$\displaystyle y=\sqrt{106} \cos(1.06) \sin (4(1.57)) + \sqrt{106} \sin(1.06) \cos (4(1.57))$

$\displaystyle y=10.29 .(0.489) \sin (6.28) + 10.29 .(0.873) \cos (6.28)$

$\displaystyle y=10.29 . (0.489) . (-0.003) + 10.29 . (0.873) . (0.99)$

$\displaystyle y=-0.0151 + 8.893$

$\displaystyle y=8.88$.....

is this correct?, would y be the answer for the amplitude?
No the amplitude is $\displaystyle \sqrt{5^2+9^2} \approx 10.3$

CB
• Nov 13th 2010, 12:39 PM
Quote:

Originally Posted by CaptainBlack
$\displaystyle y=\sqrt{5^2+9^2} \sin (4t+\alpha)$

I don't know what you think the rest means:-

No the amplitude is $\displaystyle \sqrt{5^2+9^2} \approx 10.3$

CB

I accept that the amplitude is $\displaystyle R=\sqrt{5^2+9^2} \approx 10.3$

because in the form $\displaystyle y=R Sin (4t + \alpha)=\sqrt{5^2+9^2} \sin (4t+\alpha)$
this in a sense is $\displaystyle y=Asin(t)$, where $\displaystyle A$ is the amplitude

but what does "y" mean in the question, what i have evaluated to make 8.88?
• Nov 13th 2010, 08:41 PM
CaptainBlack
Quote:

I accept that the amplitude is $\displaystyle R=\sqrt{5^2+9^2} \approx 10.3$
because in the form $\displaystyle y=R Sin (4t + \alpha)=\sqrt{5^2+9^2} \sin (4t+\alpha)$
this in a sense is $\displaystyle y=Asin(t)$, where $\displaystyle A$ is the amplitude
You are confusing the variable $\displaystyle t$ which usually represents time with the constant $\displaystyle \tau$ which represents the period (which has the same units at $\displaystyle t$)