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Math Help - need help here

  1. #1
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    need help here

    Q1) An oscillating signal voltage is given by E=125cos(wt-&)millivolts, where the angular frequency is w=123pi,phase angle is &=pi/2,and t= time in seconds. The triggering mechanism of an oscilloscope starts the sweep when E= 60mV. what is the smallest postive value of t for which the triggering occurs?

    Q2)The displacement d of a piston is given by
    d=sinwt+1/2sin2wt
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  2. #2
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    Quote Originally Posted by cheesepie
    Q1) An oscillating signal voltage is given by E=125cos(wt-&)millivolts, where the angular frequency is w=123pi,phase angle is &=pi/2,and t= time in seconds. The triggering mechanism of an oscilloscope starts the sweep when E= 60mV. what is the smallest postive value of t for which the triggering occurs?
    <br />
E(t)=125\cos(123 \pi t-\pi/2)\ \mbox{mV}<br />

    The trigger event occurs when t is the smallest positive root of:

    <br />
125\cos(123 \pi t-\pi/2)=60<br />

    Now:

    <br />
cos(\theta-\pi/2)=sin(\theta)<br />
,

    so we are interested in the roots of:

    <br />
125\sin(123 \pi t)=60<br />

    or:

    <br />
\sin(123 \pi t)=60/125=0.48<br />

    The smallest positive root is what is usual give by the arcsin function on
    your calculator so:

    <br />
123 \pi t=\arcsin(0.48)\approx 0.5<br />
,

    so:

    <br />
t\approx0.5/(123\pi) \approx 0.00129 \mbox{seconds}<br />

    RonL
    Last edited by CaptainBlack; April 26th 2006 at 10:09 AM.
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  3. #3
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    Quote Originally Posted by cheesepie

    Q2)The displacement d of a piston is given by
    d=sinwt+1/2sin2wt
    Yes but what is the question?

    RonL
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  4. #4
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    thank for ur reply


    Q2)The displacement d of a piston is given by
    d=sinwt+1/2sin2wt

    for what primary solutions of wt less than 2pi is d=o ?
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  5. #5
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    upz

    upz upz upz
    Last edited by cheesepie; April 27th 2006 at 03:10 AM.
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  6. #6
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    Quote Originally Posted by cheesepie
    thank for ur reply


    Q2)The displacement d of a piston is given by
    d=sinwt+1/2sin2wt

    for what primary solutions of wt less than 2pi is d=o ?
    Since both the terms on the RHS are 0 at wt=0 that is a solution,
    Is it the one you want though?

    RonL
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  7. #7
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    the answer is 0 , pi

    but can u show me the step ? thank alot ..
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  8. #8
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    Quote Originally Posted by cheesepie
    the answer is 0 , pi

    but can u show me the step ? thank alot ..
    <br />
d(wt)=\sin(wt)+1/2 \sin(2wt)=0<br />

    Expand the second sin function using the double angle formula:

    <br />
\sin(wt)+1/2 \sin(2wt)=\sin(wt)+\sin(wt)\cos(wt) =\sin(wt)(1+cos(wt))=0<br />

    Now the only way that \sin(wt)(1+cos(wt)) can be zero is if either
    \sin(wt)=0 or (1+cos(wt))=0, the first of these gives wt=0 \mbox{ and }\pi,
    and the second gives wt=\pi, so the solutions in [0,2\pi] are wt=0 \mbox{ and }\pi.

    RonL
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  9. #9
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    hi ...can u explain this step...not too sure ..thk..

    <br />
\sin(wt)+1/2 \sin(2wt)=\sin(wt)+\sin(wt)\cos(wt) =\sin(wt)(1+cos(wt))=0<br />
    Last edited by cheesepie; April 30th 2006 at 01:49 AM.
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  10. #10
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    upz upz upz
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  11. #11
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    Quote Originally Posted by cheesepie
    hi ...can u explain this step...not too sure ..thk..
    <br />
\sin(wt)+1/2 \sin(2wt)=\sin(wt)+\sin(wt)\cos(wt) =\sin(wt)(1+cos(wt))=0<br />
    Hello,

    as Capt.Black has suggested you should use the double angle formula of the sine function:

    \sin(2x)=2 \cdot \sin(x) \cdot \cos(x). Therefore:

    \frac{1}{2} \cdot \sin(2x)=\sin(x) \cdot \cos(x).

    So you can sustitute \frac{1}{2} \cdot \sin(2x) by \sin(x) \cdot \cos(x).

    Hope that this is of some help.

    Greetings

    EB
    .
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  12. #12
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    thanks for ur reply ..
    Last edited by cheesepie; April 30th 2006 at 08:19 AM.
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  13. #13
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    Quote Originally Posted by cheesepie
    Hi..thanks for ur reply ..i understand this step
    only dont understand sin(wt)+sin(wt)cos(wt)=sin(wt)(1+cos(wt)=0
    Hello,

    you can factorize this sum: Put the common factor (here it is sin(wt)) before the bracket. You get the contents in paranthese by dividing both summands by sin(wt). So sin(wt)/sin(wt) = 1 (not zero!!!), you'll get the sum in the bracket.

    Greetings

    EB
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  14. #14
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    Got it ..thanks ..
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