Archived Time for a trampoline jumper to lift off?

[lukasleibfried]

Homework Statement

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 x_{\max}, the maximum position of the jumper, k, the spring constant of the springs of the trampoline, m, the mass of the jumper, and g, the gravitational acceleration. Using this information, find the formula for t_{f}, the time span from when the jumper hit the mat to the time the jumper left the mat.

Homework Equations

\omega = \sqrt{\frac{k}{m}}

The Attempt at a Solution

I started by setting up the equation F = m g - k x, F being the net force and x being the position. Then, I found x_{0}, the point at which the force becomes zero, by setting up the equation m g - k x_{0} = 0. I solved the equation and found out that x_{0} = \frac{m g}{k}. Due to the fact that this system involves spring motion, it is periodic, and therefore its equation can be expressed in the form x = A \sin (\omega t) + x_{0}, A being the amplitude and \omega being the angular frequency. A = x_{\max} - x_{0}, and \omega = \sqrt{\frac{k}{m}}, and we know that F = m \frac{d^{2} x}{d t^{2}}, so F = -m \omega^{2} (x_{\max} - x_{0}) \sin (\omega t). In this equation, the y-axis is intersected at the point x_{0}, 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 - \omega t_{0} instead of \omega t_{0}. Now, we have the equation - m \omega^{2} (x_{\max} - x_{0}) \sin (- \omega t_{0}) = - m g. When we solve for t_{0}, we get t_{0} = - \frac{\sin^{-1} (\frac{g}{\omega^{2} (x_{\max} - x_{0})})}{\omega}. Now, let's solve for the time from x_{0} to x_{0}. We must find a time when - m \omega^{2} (x_{\max} - x_{0}) \sin (\omega (t_{f} - 2 |t_{0}|)) = 0. In order to make the sine factor equal to 0, thereby making the entire expression equal to zero, \omega (t_{f} - 2 |t_{0}|) = \pi, as going from x_{0} back to x_{0} is a half oscillation, as indicated by the pi. We now know that t_{f} - 2 |t_{0}| = \frac{\pi}{\omega}, meaning that t_{f} = \frac{\pi}{\omega} + 2 |t_{0}|, or 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}}}. This can be simplified to t_{f} = (2 \sin^{-1} (\frac{m g}{k x_{\max} - m g}) + \pi) \sqrt{\frac{m}{k}}.

 

[pbuk]

You have the right answer, although you could have got there a bit more easily by replacing this:

we know that F = m \frac{d^{2} x}{d t^{2}}, so F = -m \omega^{2} (x_{\max} - x_{0}) \sin (\omega t). In this equation, the y-axis is intersected at the point x_{0}, 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 - \omega t_{0} instead of \omega t_{0}. Now, we have the equation - m \omega^{2} (x_{\max} - x_{0}) \sin (- \omega t_{0}) = - m g. When we solve for t_{0}, we get t_{0} = - \frac{\sin^{-1} (\frac{g}{\omega^{2} (x_{\max} - x_{0})})}{\omega}.

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:

Now, let's solve for the time from x_{0} to x_{0}. We must find a time when - m \omega^{2} (x_{\max} - x_{0}) \sin (\omega (t_{f} - 2 |t_{0}|)) = 0. In order to make the sine factor equal to 0, thereby making the entire expression equal to zero, \omega (t_{f} - 2 |t_{0}|) = \pi, as going from x_{0} back to x_{0} is a half oscillation, as indicated by the pi.

by observing directly that that the time from $x_0$ to $x_{max}$ and back again is simply one half cycle, given by $\sqrt{\frac{m}{k}} \pi$