Trampoline fun

The formula for the period of a spring is $T=2\pi\sqrt\frac{m}{k}$ . Since we know the period is $3$ s, we can solve for $k$ in terms of $m$ , getting that $k=\frac{4\pi^2m}{9}$ . Since the forces are in equilibrium when you come to rest, the restoring force of the spring when it has been displaced below its natural height by $x$ must cancel out the force of gravity, so we have that $kx=mg$ . Plugging in our expression for $k$ and plugging in the value for acceleration due to gravity, we get that $\frac{4\pi^2m}{9}x=9.807m$ . We can cancel out the two $m$ and solve for $x$ , getting that $x \approx 2.239$ .

As a spring, the trampoline obeys Hookes' Law: that is,

$F = -kx$

where k is the spring constant and x is the compression of the spring.

We also know that your weight, mg, exerted on it is equal to the restoring force exerted by the spring since everything is at rest and the system is at equilibrium. We can therefore say that

$F$ = $mg$ = $-kx$ (Equation One)

For a simple spring, the period of oscillation T is determined by the equation

$T$ = $2{\pi}{\sqrt{m/k}}$ (Equation Two)

Changing the subject of this equation to $m/k$ ,

$m/k$ = $({T/2{\pi}})^2$ , (Equation Three)

We manipulate Equation One to get this:

${g}*m/k$ = $-x/g$ (Equation Four)

Substituting Equation Three into Equation Four:

$({g}{T/2{\pi}})^2$ = $-x$

Substituting in the appropriate we obtain the following:

$9.8*(\frac{3} {2 {\pi}})^2$ = $-x$

This gives us around x = 2.23 (the bend of the trampoline, despite being in the negative y direction, is always a positive value)

Because we are modeling the trampoline as a spring, we can say that it undergoes simple harmonic motion in its oscillations. Therefore, we know the period $T = 2 \pi \sqrt \frac {m}{k}$

We can rearrange the period equation to get $\frac {m}{k} = \frac {T^2}{4\pi^2}$

Since we are dealing with a Hookean spring (the trampoline), we know that the displacement from the natural height $x = \frac {mg}{k}$

When we combine these two equations by substituting for $x$ , we get $x = \frac {gT^2}{4\pi^2}$ or 2.23 meters.

The trampoline can be modeled as a compressed spring with constant k

Time Period= 3 seconds = 2 π t(m/k)^(1/2)

                                        => (9*k/4*π^2) = mass

Where ,m=mass Or Inertial Factor. k=Spring constant.

The trampoline is compressed by your weight = m g = (9k/π^2) g = k*x

where x is the amount the trampoline is bent.

We can drop k because it appears on both sides of the equation. Therefore:-

                =>(9*g/4*π^2) = x

Solving using g=9.8

                 => x=2.234 meters.

Firstly, we model the motion as a harmonic oscillator. Since 3 s is quite small, we have a small displacement, and the following formula holds:

$\displaystyle f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \cdots (1)$

where $f$ is the frequency of the motion, $k$ is the spring constant, and $m$ is the weight's mass, or in this case, us.

Now, we consider when we are at rest. We are now in a state of equilibrium, thus there should be no net force acting on us. There are 2 types of forces that are currently acting on us when we're at rest: the gravity and the trampoline (or spring) force.

The gravity force pulls us down onto the earth, with the magnitude of $mg$ , where $g$ being the gravitational constant.

On the other hand, since the trampoline can be modeled as a spring, we may use Hooke's law to get the magnitude of the spring's force, which is $kx$ , where $x$ is the displacement from the original position of the spring, with the direction pointing upwards.

Since we are now on rest, we have the following equality:

$\displaystyle mg = kx \Rightarrow x = \frac{m}{k} g \cdots (2)$

Now, from (1), we can rewrite the equation as

$\displaystyle \frac{m}{k} = \frac{1}{(2\pi f)^2}$

and substituting this equation to equation (2), we have:

$\displaystyle x = \frac{m}{k} g = \frac{1}{(2\pi f)^2} g$

Finally, we submit the appropriate values: assuming $g = 9.81 m/s^2$ , $\pi = 3.14$ , and from the relation $f = 1/T$ where $T$ is the period of the oscillation, we have $f = 1/(3 s) = 1/3 Hz$ . We have

$\displaystyle x = \frac{1}{(2\cdot 3.14\cdot 1/3 Hz)^2} \cdot 9.81 m/s^2 = \boxed{2.239 m}$

which is our answer.

While you're oscillating, your frequency is $\omega = \frac{2\pi}{T} = 2.093~\mbox{s}^{-1}$ . Since $\omega^2 = \frac{k}{m}$ , we can find $k = m \omega^2$ .

When you stop oscillating, the restoring elastic force is in equilibrium with your weight. Therefore, $mg = kx$ . The displacement from the equilibrium is $x = \frac{mg}{k} = \frac{g}{\omega^2} = 2.239~\mbox{m}$ .