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Math Help - amplitude of guitar string

  1. #1
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    amplitude of guitar string

    Hey everyone, i need a hand with a question i have for an assignment...

    a guitar string is vibrating in its fundamental mode, with a node at both ends. Given the length (0.390m). The max transverse acceleration at a point in the middle of the string (8900m/s^2) and max transverse velocity of 4.00 m/s

    how do i find the amplitude?

    i know the equation of a standing wave is y(x,t)= Asin(kx)sin(wt)
    where a is the amplitude of the standing wave and w denotes omega

    in the fundamental mode the wavelength is 2L...

    but im not sure where to start... can anyone point me in the right direction?
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  2. #2
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    Simple Harmonic Motion

    Hello IrrationalPI3
    Quote Originally Posted by IrrationalPI3 View Post
    Hey everyone, i need a hand with a question i have for an assignment...

    a guitar string is vibrating in its fundamental mode, with a node at both ends. Given the length (0.390m). The max transverse acceleration at a point in the middle of the string (8900m/s^2) and max transverse velocity of 4.00 m/s

    how do i find the amplitude?

    i know the equation of a standing wave is y(x,t)= Asin(kx)sin(wt)
    where a is the amplitude of the standing wave and w denotes omega

    in the fundamental mode the wavelength is 2L...

    but im not sure where to start... can anyone point me in the right direction?
    The centre point of the string will be oscillating with Simple Harmonic Motion, with equation of the form:

    x = a\sin(\omega t + \alpha)

    where x is the displacment from the equilibrium position, a is the amplitude and \omega is a constant.

    If we differentiate this equation we get

    v = \dot{x}=a\omega\cos(\omega t + \alpha)

    which gives the maximum velocity:

    • \text{vel}_{max}=a\omega

    Differentiating again, and substituting back from the original equation we get the characteristic equation of SHM:

    \ddot{x}=-\omega^2x

    So the maximum acceleration occurs when x = \pm a, and is therefore numerically

    • \text{acc}_{max}= a\omega^2

    Plug the values given in the question into these two expressions and eliminate \omega to find a.

    Grandad
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