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 $\sqrt{\frac{m}{k}} \pi$