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Archived Time for a trampoline jumper to lift off?

  1. Jan 1, 2014 #1
    1. The problem statement, all variables and given/known data

    Imagine a trampoline. Define downwards displacement as positive and upwards displacement as negative. A jumper jumps and lands on the trampoline. The given variables are [itex]x_{\max}[/itex], the maximum position of the jumper, [itex]k[/itex], the spring constant of the springs of the trampoline, [itex]m[/itex], the mass of the jumper, and [itex]g[/itex], the gravitational acceleration. Using this information, find the formula for [itex]t_{f}[/itex], the time span from when the jumper hit the mat to the time the jumper left the mat.

    2. Relevant equations

    [itex]\omega = \sqrt{\frac{k}{m}}[/itex]

    3. The attempt at a solution

    I started by setting up the equation [itex]F = m g - k x[/itex], [itex]F[/itex] being the net force and [itex]x[/itex] being the position. Then, I found [itex]x_{0}[/itex], the point at which the force becomes zero, by setting up the equation [itex]m g - k x_{0} = 0[/itex]. I solved the equation and found out that [itex]x_{0} = \frac{m g}{k}[/itex]. Due to the fact that this system involves spring motion, it is periodic, and therefore its equation can be expressed in the form [itex]x = A \sin (\omega t) + x_{0}[/itex], [itex]A[/itex] being the amplitude and [itex]\omega[/itex] being the angular frequency. [itex]A = x_{\max} - x_{0}[/itex], and [itex]\omega = \sqrt{\frac{k}{m}}[/itex], and we know that [itex]F = m \frac{d^{2} x}{d t^{2}}[/itex], so [itex]F = -m \omega^{2} (x_{\max} - x_{0}) \sin (\omega t)[/itex]. In this equation, the y-axis is intersected at the point [itex]x_{0}[/itex], not the point at which the jumper hits the mat, so we can use this to find the time difference between the original impact and point x_{0} by making the argument of the sine be [itex]- \omega t_{0}[/itex] instead of [itex]\omega t_{0}[/itex]. Now, we have the equation [itex]- m \omega^{2} (x_{\max} - x_{0}) \sin (- \omega t_{0}) = - m g[/itex]. When we solve for [itex]t_{0}[/itex], we get [itex]t_{0} = - \frac{\sin^{-1} (\frac{g}{\omega^{2} (x_{\max} - x_{0})})}{\omega}[/itex]. Now, let's solve for the time from [itex]x_{0}[/itex] to [itex]x_{0}[/itex]. We must find a time when [itex]- m \omega^{2} (x_{\max} - x_{0}) \sin (\omega (t_{f} - 2 |t_{0}|)) = 0[/itex]. In order to make the sine factor equal to 0, thereby making the entire expression equal to zero, [itex]\omega (t_{f} - 2 |t_{0}|) = \pi[/itex], as going from [itex]x_{0}[/itex] back to [itex]x_{0}[/itex] is a half oscillation, as indicated by the pi. We now know that [itex]t_{f} - 2 |t_{0}| = \frac{\pi}{\omega}[/itex], meaning that [itex]t_{f} = \frac{\pi}{\omega} + 2 |t_{0}|[/itex], or [itex]t_{f} = \frac{\pi}{\sqrt{\frac{k}{m}}} + 2 \frac{\sin^{-1} (\frac{g}{\frac{k}{m} (x_{\max} - \frac{mg}{k})})}{\sqrt{\frac{k}{m}}}[/itex]. This can be simplified to [itex]t_{f} = (2 \sin^{-1} (\frac{m g}{k x_{\max} - m g}) + \pi) \sqrt{\frac{m}{k}}[/itex].
     
  2. jcsd
  3. Feb 23, 2016 #2
    You have the right answer, although you could have got there a bit more easily by replacing this:
    by noting that the amplitude of the oscillation is ## (x_{max} - x_0) ## and so ## x_0 ## is given by ## x_0 = (x_{max} - x_0) \sin (\omega t_0) ## from which you quickly get ## t_0 = \sqrt{\frac{m}{k}} \sin^{-1} (\frac{m g}{k x_{\max} - m g}) ##, and replacing this:
    by observing directly that that the time from ## x_0 ## to ## x_{max} ## and back again is simply one half cycle, given by ## \sqrt{\frac{m}{k}} \pi ##
     
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