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匿名 發問於 科學及數學其他 - 科學 · 9 年前

天同><! (SHM)

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For SHM

有齊

x=Acos(ωt)

x=Asin(ωt+θ)

我知cos(x) = sin(x+(π/2))

不過2條式d出尼既v同a會有唔同

例如話我d個cosx會出-sinx

d個sinx會出個cosx

而θ係個constant

d出尼就會多左個負號既分別

主要係想問 其實2條式個分別係邊度同埋幾時用sin幾時用cos?

更新:

意思就係話... general 就係Asin(ωt) 而in diff. case, 我地可以derive Acos(ωt) = =?

更新 2:

thanks a lot: )

there maybe still a lot of questions to seek help from you = =...~

1 個解答

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  • 天同
    Lv 7
    9 年前
    最愛解答

    The simple answer is: which equation to be used depends on the initial condition of the system. That is, the displacement (value of x) at t = 0.

    The equation: x=Asin(ωt+θ) is a general equation for a simple harmonic motion (SHM). It suits all conditions. If we take a simple pendulumn for example, when the pendulumn bob is displaced to the right most extreme position and then released, the initial condition is thus t=0, x = A. Substitute these values into the equation, we get, A=Asin(θ). This gives θ = pi/2 radians.

    The equation for such system is therefore represented by the equation:

    x = Asin(ωt + pi/2) which is x =A.cos(ωt)

    Assume the pendulum bob is at the equilibrium position (x=0) at t=0 (these are the initial condition for such system) moving towards the right, when we put x = 0, t=0 into the general equation, we get: 0 =Asin(θ), i.e. θ = 0. Hence, such system follows the equation: x =A.sin(ωt)

    Becuase the two systems described above are of different initial conditions, it is apparent that they have different velocities (and acceleration too) at different time. For example, after a quarter of a period, the first system will be at the equilibrium position and have the max velocity. The velocity direction during this first quarter period is towards the left, hence -ve in sign. For the second system, it will be at the extreme position and have zero velocity.

    As a further example, if the pendulumn bob is first displaced to the left most extreme position and released, then the initial conditions are t=0, x = -A. The general equation now becomes: -A=A.sin((θ), i.e. θ = -pi/2

    The equation of this system is: x = A.sin(ωt -pi/2) = A.sin[-(pi/2-ωt)] = -A.sin(pi/2-(ωt) = -A.cos(ωt)

    2012-03-07 21:15:17 補充:

    Please refer to further explanation on the velocity equations in the "opinion(意見)" section.

    2012-03-07 21:20:53 補充:

    The general equation x=A.sin(wt+pi/2) is equivalent to x=Acos(wt)

    Differentiate the general equation for velocity v

    v = A.d[sin(wt+pi/2)]dt = Aw.cos(wt+pi/2) = -Aw.sin(wt)

    2012-03-07 21:23:25 補充:

    This is the same as obtained by differentiating the special equation x = A.cos(wt)

    Hence, no matter which equation (general form or special form) you use, you should get the same result for velocity and acceleration.

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