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Math Help - Reciprocating piston problem, can you solve it?

  1. #1
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    Reciprocating piston problem, can you solve it?

    The displacement, in metres, of a reciprocating piston, from its point of equilibrium (the mean) is given by the following function:


    x= 3 cos( 4t +
    π
    /2) where t is time in seconds.


    Starting from rest, find the first three times at which the piston passes through the mean position at x=0
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  2. #2
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    Re: Reciprocating piston problem, can you solve it?

    What are you having trouble with? Do you know for what values of $\theta$, $\cos \theta = 0$?
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  3. #3
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    Re: Reciprocating piston problem, can you solve it?

    I don't know where to start with the question and no I don't know for what values of theta or cos theta=o
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  4. #4
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    Re: Reciprocating piston problem, can you solve it?

    Quote Originally Posted by TehSwedishBeast View Post
    I don't know where to start with the question and no I don't know for what values of theta or cos theta=o
    your told that the displacement $x$ is given by

    $x= 3 \cos(4t + \dfrac{\pi}{2})$

    and you need to find when $x=0$ so just set this expression equal to zero and solve it.

    $0= 3 \cos(4t + \dfrac{\pi}{2})$

    $0=\cos(4t + \dfrac{\pi}{2})$

    Now you have to know that $\cos(\theta)=0$ when $\theta=\pi (n+\dfrac{1}{2})$ for n being some integer. Set n=0,1,2 to find the first 3 times $x=0$

    so for example with n=0

    $4t + \dfrac{\pi}{2} = \dfrac{\pi}{2}$

    $4t=0$

    $t=0$

    you can solve for the t's corresponding to n=1 and n=2
    Thanks from TehSwedishBeast
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  5. #5
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    Re: Reciprocating piston problem, can you solve it?

    Quote Originally Posted by TehSwedishBeast View Post
    I don't know where to start with the question and no I don't know for what values of theta or cos theta=o
    romsek gave you the answer, but the only way you would not know for what values of $\theta$ give $\cos \theta = 0$ is if you never studied trigonometry. I would recommend holding off on answering the question until after you study trigonometry.
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  6. #6
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    Re: Reciprocating piston problem, can you solve it?

    I understand now so for n=1 would it be:
    4t + π/2= π/2
    4t= 1
    t= 0.25

    and for n=2 would it be:
    4t + π/2 = π/2
    4t= 2
    t=0.5

    Am I right with that?
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  7. #7
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    Re: Reciprocating piston problem, can you solve it?

    Quote Originally Posted by TehSwedishBeast View Post
    I understand now so for n=1 would it be:
    4t + π/2= π/2
    4t= 1
    t= 0.25

    and for n=2 would it be:
    4t + π/2 = π/2
    4t= 2
    t=0.5

    Am I right with that?
    no.

    for n=1 we have

    $4t + \dfrac{\pi}{2}=\pi(1+\frac{1}{2})=\dfrac{3\pi}{2}$

    $4t = \pi$

    $t=\dfrac{\pi}{4}$

    for n=2 the right hand side is $\pi(2+\frac{1}{2})=\dfrac{5\pi}{2}$

    Now solve this for t.
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