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Math Help - Finding the amplitude

  1. #1
    Junior Member madmax29's Avatar
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    Finding the amplitude

    I have

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

    angular freq, w= 4 rads,

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

    how to you find the amplitude???
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by madmax29 View Post
    I have

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

    angular freq, w= 4 rads,

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

    how to you find the amplitude???
    Trigonometry!

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

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

    etc..

    CB
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  3. #3
    Junior Member madmax29's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Trigonometry!

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

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

    etc..

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

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

    Now;

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

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

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

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

    y=-0.0151 + 8.893

    y=8.88.....

    is this correct?, would y be the answer for the amplitude?
    Last edited by madmax29; November 13th 2010 at 12:33 PM. Reason: changed degs to radians
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  4. #4
    Junior Member madmax29's Avatar
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    Found this;


    where y from the question = Ao on the diagram above

    x-axis would be time (in secs)
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by madmax29 View Post
    Therefore; \alpha=\arctan(9/5)= \arctan(1.8)= 60.94 in degrees, then:

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

    Now;

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

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

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

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

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

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

    y=-0.0151 + 8.893

    y=8.88.....

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

    CB
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  6. #6
    Junior Member madmax29's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    y=\sqrt{5^2+9^2} \sin (4t+\alpha)

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



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

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

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

    but what does "y" mean in the question, what i have evaluated to make 8.88?
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by madmax29 View Post
    I accept that the amplitude is R=\sqrt{5^2+9^2} \approx 10.3

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

    but what does "y" mean in the question, what i have evaluated to make 8.88?
    You are confusing the variable t which usually represents time with the constant \tau which represents the period (which has the same units at t)

    CB
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