# simple harmonic motion

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• Jan 3rd 2008, 07:37 AM
franklina
simple harmonic motion
a bungee jumper falls a total distance of 48 metres, of which the first 20 metres are free fall under gravity alone before the string becomes taut. find the total time she takes to the lowest point of her jump, and the greatest speed she attains.

please help
• Jan 3rd 2008, 12:08 PM
topsquark
Quote:

Originally Posted by franklina
a bungee jumper falls a total distance of 48 metres, of which the first 20 metres are free fall under gravity alone before the string becomes taut. find the total time she takes to the lowest point of her jump, and the greatest speed she attains.

please help

She falls 20 m before the bungee starts to slow her down, so she takes
$t = \sqrt{\frac{2d}{g}} = 2.02031~s$ to fall this distance. The bungee slows her down at this point, so the fastest she is moving is $v = \sqrt{2gd} = 19.799~m/s$. (I can show you how do derive these formulas if you like.)

Now, she starts out at 20 m with her highest speed and ends at 48 m with a speed of 0. This is going to represent 1/4 of the cycle of a cosine function. This cosine function takes the form:
$y = y_0 + A~cos(\omega t)$
where $y_0 = 20~m$, A is the amplitude of the "wave" which is 48 m - 20 m = 28 m, and $\omega$ is the angular frequency of the oscillation of her motion on the bungee. The amount of time it takes for one period of the motion is given by $\omega T = 2 \pi$, or
$T = \frac{2 \pi}{\omega}$

Now, $\omega$ is given by the spring equation by
$\omega = \sqrt{\frac{k}{m}}$
where k is the spring constant of the bungee cord and m is the mass of the diver. As we are given neither of these data we cannot finish the problem.

We can, of course, give it in its form depending on k and m, though. The time it takes for the diver to fall from where the bungee starts to retard her motion to her lowest point is
$t = \frac{T}{4} = \frac{\pi}{2}\sqrt{\frac{m}{k}}$

So the total time for the fall will be
$2.02031 + \frac{\pi}{2}\sqrt{\frac{m}{k}}$ seconds.

-Dan
• Jan 3rd 2008, 02:29 PM
franklina
shm question
thank you very much for attempting the question.

when the string becomes taut, i think that this just means that its tension is greater than 0, i.e. there is still time while the resultant force is downward and thus the speed is increasing
your answer to the second question is based on assumption that the equilibrium position is at 20m rather than somewhere between 20 and 48.
• Jan 3rd 2008, 03:07 PM
JaneBennet
You don’t need to calculate the tension in the string! You just need to calculate the deceleration for the second part of her jump.

The first part is straight forward. Use $s=ut+\frac{1}{2}\,at^2$ to find the time of her freefall: $20=0t+\frac{1}{2}\,(10)t^2\ \Rightarrow\ t=2$ (taking $g=10\ \textrm{m}\,\textrm{s}^{-2}$). Then use $v=u+at$ to find her speed at the end of the freefall (this will also be the greatest speed she attains): $v=0+(10)(2)=20\ \textrm{m}\,\textrm{s}^{-1}$.

For the second part, first find her deceleration using $v^2=u^2+2as$: $0^2=20^2+2a(48-20)\ \Rightarrow\ a=-\frac{50}{7}$. Now use $s=ut+\frac{1}{2}\,at^2$ again to find the time for the second part of her descent: $28=20t+\frac{1}{2}\cdot\left(-\frac{50}{7}\right)t^2$. This can be arranged into the quadratic equation $25t^2-140t+196=0\ \Rightarrow\ (5t-14)^2=0\ \Rightarrow\ t=2.8$.

Hence the total time of her fall is 2 + 2.8 = 4.8 seconds.
• Jan 3rd 2008, 04:02 PM
topsquark
Quote:

Originally Posted by JaneBennet
For the second part, first find her deceleration using $v^2=u^2+2as$

Unfortunately this assumes that her deceleration is constant. A bungee acts like a spring in that the tension is proportional to the distance the bungee is stretched, so
$a \propto x$
meaning acceleration is not constant.

-Dan
• Jan 3rd 2008, 04:27 PM
topsquark
Quote:

Originally Posted by franklina
thank you very much for attempting the question.

when the string becomes taut, i think that this just means that its tension is greater than 0, i.e. there is still time while the resultant force is downward and thus the speed is increasing
your answer to the second question is based on assumption that the equilibrium position is at 20m rather than somewhere between 20 and 48.

The speed of the diver can't be increasing after the bungee starts to stretch because the force the bungee exerts on the diver is upward, whereas the velocity of the diver is downward.

Yes, 20 m is the equilibrium height since this is the point on the oscillation that has the greatest speed.

-Dan
• Jan 4th 2008, 01:49 AM
franklina
unfortunately, again yu are wrong,
though you are correct that there is a force (tension) acting in the opposite direction to the string, since tension is proportional to displacement from where the tension starts to act (in this case 20m) there must be a point at which the displacement is small enough that the tension will be smaller than the weight force,

therefore, when the string becomes taut, for some length of time afterward, the jumper will still be accelerating downwards (albeit with a decreasing acceleration).
• Jan 4th 2008, 04:40 AM
topsquark
Quote:

Originally Posted by franklina
unfortunately, again yu are wrong,
though you are correct that there is a force (tension) acting in the opposite direction to the string, since tension is proportional to displacement from where the tension starts to act (in this case 20m) there must be a point at which the displacement is small enough that the tension will be smaller than the weight force,

therefore, when the string becomes taut, for some length of time afterward, the jumper will still be accelerating downwards (albeit with a decreasing acceleration).

You are correct. Unfortunately I can't finish my post as the baby is crying. I'll get back to you.

-Dan
• Jan 4th 2008, 05:36 AM
franklina
i think that the answer should be purely numerical, as, with other questions in the book, if the answers are algebraic the question says so.
but dont worry too much about it.
• Jan 4th 2008, 08:32 AM
topsquark
Quote:

Originally Posted by franklina
i think that the answer should be purely numerical, as, with other questions in the book, if the answers are algebraic the question says so.
but dont worry too much about it.

You are indeed right that this can be solved without explicit values for m and k. But to find the solution isn't pretty.

My problem earlier was that I was trying to simplify it to shortcut the work. This is not always a good idea and gave me incorrect answers. So without further ado, here is the solution start to finish. (Be warned, it's long.)

I am going to start by labeling a coordinate system. I'm going to put the origin as the "ground" and I'm going to label upward as positive. So the diver starts at y(0) = 48 m.

The problem goes in two stages: freefall to a height of 28 m, then the bungee starts to affect the motion and the diver falls the additional 28 m.

The freefall part of the motion is simple and gives a time of fall of
$t = \sqrt{2 \cdot 20g} \approx 2.02031~s$
and a velocity at the end of the 20 m fall of
$v = -\sqrt{2 \cdot 20 \cdot g} \approx -19.799~m/s$

Now for the hard part.
I'm going to let c = 28 m and d = 20 m. Since we don't know a value for m and k yet, I'm just going to leave these as m and k for now.

Newton's 2nd Law says:
$\sum F = ma = -mg + T = -mg + k(c - y)$

So we need to solve
$m\frac{d^2y}{dt^2} + ky = kc - mg$
with the initial conditions $y(0) = c$ and $v(0) = -\sqrt{2dg}$.

After a little work I find the solution to be
$y(t) = -\sqrt{2dg \cdot \frac{m}{k} }~sin \left ( t \sqrt{\frac{k}{m}} \right ) + \left ( g \cdot \frac{m}{k} \right )~cos \left ( t \sqrt{\frac{k}{m}} \right ) + \left ( c - g \cdot \frac{m}{k} \right )$

To simplify this a bit I'm going to define $\omega = \sqrt{\frac{k}{m}}$, so
$y(t) = -\sqrt{\frac{2dg}{\omega ^2}}~sin( \omega t) + \left ( \frac{g}{\omega ^2} \right )~cos( \omega t) + \left ( c - \frac{g}{\omega ^2} \right )$

So
$v(t) = -\sqrt{2dg}~cos( \omega t) - \left ( \frac{g}{\omega } \right )~sin( \omega t)$
and
$a(t) = \omega \sqrt{2dg}~sin( \omega t) - g~cos( \omega t)$

We want to first get the time of fall to the lowest point, which is 0 m. So we need to solve v(t) = 0 for t:
$v(t) = -\sqrt{2dg}~cos( \omega t) - \left ( \frac{g}{\omega } \right )~sin( \omega t) = 0$

This gives
$t = \frac{1}{\omega } tan^{-1} \left ( -\omega \sqrt{\frac{2d}{g}} \right )$

Now, if you plug the numbers in (assuming reasonable values for k and m for the moment) you will find that t is negative. This actually corresponds to a maximum of the y(t) function in a position that doesn't physically exist. To find the minimum we shift the result of the inverse tangent function by $\pi$. So we get
$t = \frac{1}{\omega } tan^{-1} \left ( -\omega \sqrt{\frac{2d}{g}} \right ) + \frac{\pi}{\omega }$

We don't have any numbers yet to calculate t. But we do have the condition that y(t) = 0 m at this time for the minimum:
$y_{min} = -\sqrt{\frac{2dg}{\omega ^2}}~sin \left ( \omega
\left [ \frac{1}{\omega } tan^{-1} \left ( -\omega \sqrt{\frac{2d}{g}} \right ) + \frac{\pi}{\omega } \right ] \right ) +$
$\left ( \frac{g}{\omega ^2} \right )~cos \left ( \omega
\left [ \frac{1}{\omega } tan^{-1} \left ( -\omega \sqrt{\frac{2d}{g}} \right ) + \frac{\pi}{\omega } \right ] \right ) + \left ( c - \frac{g}{\omega ^2} \right ) = 0$

I get that
$-\sqrt{\frac{2dg}{\omega ^2}} \left ( \frac{\omega \sqrt{2d}}{\sqrt{2d \omega ^2 + g}} \right ) - \left ( \frac{g}{\omega ^2} \right ) \left ( \frac{\sqrt{g}}{\sqrt{2d \omega ^2}} \right ) + \left ( c - \frac{g}{\omega ^2} \right ) = 0$

This can be solved numerically for $\omega$ and I get that $\omega \approx 1.19482~Hz$.

Now we can find the time of fall from the point where the freefall ends:
$t = \frac{1}{\omega } tan^{-1} \left ( -\omega \sqrt{\frac{2d}{g}} \right ) + \frac{\pi}{\omega } \approx 1.64337~s$
so the total time of fall is
$t = 2.02031~s + 1.64337~s = 3.66368~s$.

To find the maximum speed we use a similar process. We need to find the minimum of v(t). (Recall that v(t) is negative for the time period we are talking about because the diver is falling. So a minimum velocity will correspond to the maximum speed.)
$a(t) = \omega \sqrt{2dg}~sin( \omega t) - g~cos( \omega t) = 0$

I get
$t = \frac{1}{\omega} tan^{-1} \left ( \frac{g}{\omega \sqrt{2dg}} \right ) \approx 0.328705~s$

So the maximum speed will be
$v_{max} = \sqrt{2dg}~cos( \omega t) + \left ( \frac{g}{\omega } \right )~sin( \omega t) \approx 21.4307~m/s$

Did your teacher really expect you to go through all that?

-Dan
• Jan 4th 2008, 12:09 PM
franklina
i really am very grateful for the effort you put into this!
thanks again!