# Time for a trampoline jumper to lift off?

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• lukasleibfried
In summary, the formula for t_{f}, the time span from when the jumper hit the mat to the time the jumper left the mat, is given by t_{f} = (2 \sin^{-1} (\frac{m g}{k x_{max} - m g}) + \pi) \sqrt{\frac{m}{k}}.
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 $t_{f} = (2 \sin^{-1} (\frac{m g}{k x_{\max} - m g}) + \pi) \sqrt{\frac{m}{k}}$.

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

## 1. How does a trampoline help a jumper lift off?

A trampoline uses the elasticity of its surface to propel a jumper upwards. When the jumper applies force to the trampoline, the surface deforms and stores potential energy. This potential energy is released as the surface bounces back, pushing the jumper upwards.

## 2. What factors affect the time it takes for a trampoline jumper to lift off?

The time it takes for a trampoline jumper to lift off is affected by several factors, including the weight and height of the jumper, the stiffness and shape of the trampoline, and the force applied by the jumper. Additionally, air resistance and external forces like wind can also impact the time.

## 3. How can a jumper increase their lift off time on a trampoline?

A jumper can increase their lift off time on a trampoline by increasing the force they apply to the surface, as well as their speed and height while jumping. They can also choose a trampoline with a stiffer surface or a larger surface area, which can provide more potential energy for a higher lift off.

## 4. Does the location of the trampoline affect the lift off time?

Yes, the location of the trampoline can affect the lift off time. Factors such as altitude and air density can impact the resistance and force applied to the jumper, which can affect the time it takes to lift off. Additionally, the surface on which the trampoline is placed can also affect the bounce and therefore the lift off time.

## 5. Are there any safety precautions to consider when using a trampoline?

Yes, there are several safety precautions to consider when using a trampoline. These include proper supervision, ensuring the trampoline is in good condition, and following proper jumping techniques. It is also important to only use a trampoline with proper safety features, such as a safety net and padding around the edges.

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