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Math Help - Trigonometric Identity Problem

  1. #1
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    Trigonometric Identity Problem

    Hi, I'm aditya. Nice to meet you. I have a problem with vibration derivation.
    I don't know the answer,

    How could possibly the trigonometric function of



    become the form



    Any chance that the former can be the latter form? I've searched and looked thorough and still don't have the answer until now.

    The list of what I've been through is
    List of trigonometric identities - Wikipedia, the free encyclopedia
    Abramowitz and Stegun, p. 80, 4.4.42
    Stumbled upon my own hand-held (writing) derivation which was not satisfactory.

    I just don't know how to solve this problem. I think it's a mathematics one. So i tried to submit the problem via this math Forum in there is possibility of someone solving "the unsolved".

    It was originally a physics problem. The Vibration derivation of This problem is as follows (it is from the book of Singiresu Rao, Mechanical Vibrations).

    Thanks. Any Help would be very appreciated
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  2. #2
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    Hello aditya

    Welcome to Math Help Forum!
    Quote Originally Posted by aditya View Post
    Hi, I'm aditya. Nice to meet you. I have a problem with vibration derivation.
    I don't know the answer,

    How could possibly the trigonometric function of



    become the form



    Any chance that the former can be the latter form? I've searched and looked thorough and still don't have the answer until now.

    The list of what I've been through is
    List of trigonometric identities - Wikipedia, the free encyclopedia
    Abramowitz and Stegun, p. 80, 4.4.42
    Stumbled upon my own hand-held (writing) derivation which was not satisfactory.

    I just don't know how to solve this problem. I think it's a mathematics one. So i tried to submit the problem via this math Forum in there is possibility of someone solving "the unsolved".

    It was originally a physics problem. The Vibration derivation of This problem is as follows (it is from the book of Singiresu Rao, Mechanical Vibrations).

    Thanks. Any Help would be very appreciated
    The answer is that, yes, we can write a\sin x +b\cos x as \sqrt{a^2+b^2}\sin(x+\phi), but, no, this is not the same as a\sin x +b\cos x =\sqrt{a^2+b^2}. It is clearly impossible that it should be so, since \sqrt{a^2+b^2} is a constant, whereas a\sin x +b\cos x is a variable function of x.

    The first identity is arrived at like this:

    Let a\sin x +b\cos x= r\sin(x+\phi), for some values of r and \phi

    Then using the addition formula \sin(A+B)=\sin A\cos B +\cos A\sin B,

    a\sin x +b\cos x= r(\sin x\cos\phi+\cos x\sin\phi)

    Compare coefficients of \sin x and \cos x:

    a = r\cos\phi (1)

    b=r\sin\phi (2)

    Square (1) and (2) and add:

    a^2 +b^2=r^2\cos^2\phi+r^2\sin^2\phi=r^2, since \cos^2\phi+\sin^2\phi=1

    \Rightarrow r = \sqrt{a^2+b^2}

    Divide (2) by (1):

    \tan\phi = \frac{b}{a}

    So a\sin x +b\cos x= r\sin(x+\phi), where r = \sqrt{a^2+b^2} and \tan\phi = \frac{b}{a}

    Does that help to clear things up?

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello aditya

    Welcome to Math Help Forum!
    The answer is that, yes, we can write a\sin x +b\cos x as \sqrt{a^2+b^2}\sin(x+\phi), but, no, this is not the same as a\sin x +b\cos x =\sqrt{a^2+b^2}. It is clearly impossible that it should be so, since \sqrt{a^2+b^2} is a constant, whereas a\sin x +b\cos x is a variable function of x.

    The first identity is arrived at like this:

    Let a\sin x +b\cos x= r\sin(x+\phi), for some values of r and \phi

    Then using the addition formula \sin(A+B)=\sin A\cos B +\cos A\sin B,

    a\sin x +b\cos x= r(\sin x\cos\phi+\cos x\sin\phi)

    Compare coefficients of \sin x and \cos x:

    a = r\cos\phi (1)

    b=r\sin\phi (2)

    Square (1) and (2) and add:

    a^2 +b^2=r^2\cos^2\phi+r^2\sin^2\phi=r^2, since \cos^2\phi+\sin^2\phi=1

    \Rightarrow r = \sqrt{a^2+b^2}

    Divide (2) by (1):

    \tan\phi = \frac{b}{a}

    So a\sin x +b\cos x= r\sin(x+\phi), where r = \sqrt{a^2+b^2} and \tan\phi = \frac{b}{a}

    Does that help to clear things up?

    Grandad
    Sorry took long to reply..
    I have thought o f it too. But the derivation from the book
    Make me a little Dizzy..
    Mechanical Vibration Singiresu Rao..



    The Ampitude Response i think is from 6.. and



    and the Phase Difference i think is from 7..



    The complete derivation of the equation is as follows in the attachments.
    Attached Files Attached Files
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  4. #4
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    Forced, Damped Harmonic Motion

    Hello aditya
    Quote Originally Posted by aditya View Post
    Sorry took long to reply..
    I have thought o f it too. But the derivation from the book
    Make me a little Dizzy..
    Mechanical Vibration Singiresu Rao..



    The Ampitude Response i think is from 6.. and



    and the Phase Difference i think is from 7..



    The complete derivation of the equation is as follows in the attachments.
    There are, sadly, three mistakes in the text that you are trying to understand.

    The differential equation representing forced, damped harmonic motion is correct:

    m\ddot{x} + c\dot{x}+kx = F_0\cos \omega t (1)

    The Particular Integral can then be written as x(t)=X\cos(\omega t - \phi) (2) Correct.

    If we differentiate this twice, we get:

    \dot{x} = -X\omega\sin(\omega t - \phi)

    \ddot{x} = -X\omega^2\cos(\omega t -\phi)

    Substituting these into equation (1), and re-arrange:

    X[(k-m\omega^2)\cos(\omega t - \phi)-c\omega\sin(\omega t - \phi)]=F_0\cos\omega t (3) Correct.

    We now use the trig addition formulae to expand \cos(\omega t - \phi) and \sin(\omega t - \phi) and then compare the coefficients of \cos \omega t and \sin \omega t on both sides of the equation, in much the same way as in my first posting.

    The coefficient of \cos\omega t on the RHS is F_0. This gives:

    X[(k-m\omega^2)\cos\phi +c\omega\sin\phi] = F_0 (6) Correct.

    The coefficient of \sin\omega t on the RHS is 0. This gives:

    X[(k-m\omega^2)\sin\phi - c\omega\cos\phi]=0 (7). There are two errors in the printed text of this equation!

    Equations (6) and (7) are now of the form:

    X[a\cos\phi+b\sin\phi]=F_0

    a\sin\phi-b\cos\phi=0

    where a = (k-m\omega^2) and b = c\omega

    From which we get:

    \tan\phi =\frac{\sin\phi}{\cos\phi}= \frac{b}{a}

    \Rightarrow \sin\phi = \frac{b}{\sqrt{a^2+b^2}}, \, \cos\phi= \frac{a}{\sqrt{a^2+b^2}}

    \Rightarrow X\Big(\frac{a^2}{\sqrt{a^2+b^2}}+\frac{b^2}{\sqrt{  a^2+b^2}}\Big) = F_0

    \Rightarrow X\Big(\frac{a^2+b^2}{\sqrt{a^2+b^2}}\Big) = F_0

    \Rightarrow X = \frac{F_0}{\sqrt{a^2+b^2}} = \frac{F_0}{\sqrt{(k-m\omega^2)^2+c^2\omega^2}} (8), not as printed. (Can you see the error?)

    and \phi=\tan^{-1}\Big(\frac{c\omega}{k-mw^2}\Big) (9) Correct!

    This, then, is the Particular Solution to the Differential Equation. You will also need to consider the solution to the Complementary Function m\ddot{x} + c\dot{x}+kx = 0 in order to get the General Solution.

    Grandad
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