Question

A $\mathbf{65}.\mathbf{0}-\mathbf{kg}$bungee jumper steps off a bridge with a light bungee cord tied to her body and to the bridge. The unstretched length of the cord isrole="math" localid="1660198723821" $\mathbf{11}.\mathbf{0} \mathbf{m}$ . The jumper reaches the bottom of her motion role="math" localid="1660198746464" $\mathbf{36}.\mathbf{0} \mathbf{m}$below the bridge before bouncing back. We wish to find the time interval between her leaving the bridge and her arriving at the bottom of her motion. Her overall motion can be separated into an role="math" localid="1660198763301" $\mathbf{11}.\mathbf{0}-\mathbf{m}$free fall and a role="math" localid="1660198776457" $\mathbf{25}.\mathbf{0}-\mathbf{m}$section of simple harmonic oscillation. (a) For the free-fall part, what is the appropriate analysis model to describe her motion? (b) For what time interval is she in free fall? (c) For the simple harmonic oscillation part of the plunge, is the system of the bungee jumper, the spring, and the Earth isolated or non- isolated? (d) From your response in part (c) find the spring constant of the bungee cord. (e) What is the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper? (f) What is the angular frequency of the oscillation? (g) What time interval is required for the cord to stretch by role="math" localid="1660198795390" $\mathbf{25}.\mathbf{0} \mathbf{m}$? (h) What is the total time interval for the entirerole="math" localid="1660198809416" $\mathbf{36}.\mathbf{0}-\mathbf{m}$ drop?

Short answer

(a) Constant acceleration of $9.8 {\text{m/s}}^2$

(b)$1.5 \text{s}$

(c) System is isolated

(d)$73.38 \text{N/m}$

(e)$19.68 \text{m}$

(f)$1.06 \text{rad/s}$

(g)$2.01 \text{s}$

(h)$3.51 \text{s}$

Step-by-step solution

Identification of given data

Mass of bungee jumper is$m=65 \text{kg}$

The unstretched length of the cord is$y_i=11.0 \mathrm{m}$

The jumper reaches the bottom of her motion$h=36.0 \mathrm{m}$ below the bridge before bouncing back

Distance of free fall is$y_{\text{ff}}=11.0 \mathrm{m}$

Distance of simple harmonic oscillation is$y_{\text{shm}}=25 \text{m}$

Significance of free fall

When a body is just being moved by the gravity of the earth, such situation is referred to as free fall. The motion of the ball will be accelerated because an outside force is operating on it. The term "acceleration due to gravity" is also used to describe this free-fall speed.

(a) Determining the appropriate analysis model for free-fall

When a body falls under free fall it is in the influence of gravitational field of earth only. So that her motion will be constant acceleration of$9.8 {\text{m/s}}^2$

(b) Determining the time interval when she is in free fall

To determine the time interval use the following formula for free fall

$y_{ff}=ut+\frac{1}{2}g{t}^2$

Where,$u$is the initial velocity and it is taken as zero

Substitute all the values in the above equation

$\begin{array}{rcl}11 \text{m}& =& \left(0\right)t+\frac{1}{2}\times 9.8 {\text{m/s}}^{\text{2}}\times t\\ t& =& \sqrt{\frac{2\times 11 \text{m}}{9.8 {\text{m/s}}^{\text{2}}}}\\ & =& 1.5 \text{s}\end{array}$

Hence the time interval when she is in free fall is$1.5 \text{s}$

(c) Determining that whether the system is isolated or non- isolated

The system consisting of the spring, the bungee jumper, and the earth is an isolated system based on the particle in simple harmonic motion notion.

(d) Determining the spring constant of the bungee cord

Energy is conserved within an isolated system. Take the first point where she steps from the bridge and the last point at the bottom of her motion.

${\left(K+U_g+U_e\right)}_i={\left(K+U_g+U_e\right)}_f$

Where,$K$is the kinetic energy,$U_g$is the gravitational potential energy and $U_e$is the elastic potential energy

Substitute all the values in above equation

$\begin{array}{rcl}{\left(0+mgh+0\right)}_i& =& {\left(0+0+\frac{1}{2}k{y_{\text{shm}}}^2\right)}_f\\ \left(65 \text{kg}\right)\times \left(\text{9.8} {\text{m/s}}^2\right)\times \left(36 \text{m}\right)& =& \frac{1}{2}\times k\times {\left(25 \text{m}\right)}^2\\ k& =& 73.38 \text{N/m}\end{array}$

Hence the spring constant of the bungee cord is$73.38 \text{N/m}$

(e) Determining the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper

The extension of spring at equilibrium is,

$\begin{array}{rcl}x& =& \frac{F}{k}\\ & =& \frac{mg}{k}\end{array}$

Substitute all the values in the above equation

$\begin{array}{rcl}dx& =& \frac{\left(65 \text{kg}\right)\times \left(9.8 {\text{m/s}}^{\text{2}}\right)}{73.38 \text{N/m}}\\ & =& 8.68 \text{m}\end{array}$

So this point is $8.68 \text{m}+11 \text{m}=19.68 \text{m}$ below the bridge and the amplitude of her oscillation is$36 \text{m}-19.68 \text{m}=16.32 \text{m}$

(f) Determining the angular frequency of the oscillation

Formula for angular frequency of the oscillation is expressed as

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

Substitute all the values in above equation

$$\begin{gathered} \omega =\sqrt{\frac{73.38 \text{N/m}}{65 \text{kg}}} \\ =1.06 \text{rad/s} \end{gathered}$$

(g) Determining the time interval is required for the cord to stretch by 25.0 m 

Set the spring so that the bungee jumper is in its balanced position at $x=0$. The lowest portion of the drop corresponds to $x=+16.32 \text{m}$(we have taken down as positive) and is where the spring starts to stretch at $x=-8.68 \text{m}$position. Consider the phase to be zero at the greatest downward extension $x=+16.32 \text{m}$.

We find that the phase($\omega t$) was $25 \text{m}$higher where $x=-8.68 \text{m}$(above the equilibrium point):

$x=A\cos \omega t$

At $t=0$

$\begin{array}{rcl}x& =& 16.32 \text{m}\cos 0^0\\ & =& 16.32 \text{m}\end{array}$

Where $x=-8.68 \text{m}$

$\begin{array}{rcl}\left(-8.68 \text{m}\right)& =& \left(16.32 \text{m}\right)\cos \omega t\\ \omega t& =& \pm {122}^0 \text{or} \pm 2.13 \text{rad}\end{array}$

From the below equation get an idea of choosing the positive or negative sign for $\omega t$

$\begin{array}{rcl}dv& =& \frac{dx}{dt}\\ & =& -\omega A\sin \omega t\end{array}$

At $x=-8.68 \text{m}$, $v$is in downward, which means by our choice of positive direction, $v$is positive

So the take $\omega t= -2.13 \text{rad}$

$\begin{array}{rcl}dv& =& -\omega A\sin \left(-2.13 \text{rad}\right)\\ & =& +\left(0.848\right)\omega A\end{array}$

Therefore

$\begin{array}{rcl}\omega t& =& 1.06t\\ & =& -2.13 \text{rad}\\ t& =& \frac{-2.13 \text{rad}}{1.06 \text{rad/s} }\\ & =& -2.01 \text{s}\end{array}$

Meaning of $t=-2.01 \text{s}$ is that when the spring begins to stretch and$t=0$ when the jumper reaches the bottom of the jump then$t=+2.01 \text{s}$ is the time over which the spring stretches.

(h) Determining the total time interval for the entire 36.0-m drop

So the total time interval for the entire $36.0-\mathrm{m}$drop is,

$\begin{array}{rcl}T& =& 1.5 \text{s}+2.01 \text{s}\\ & =& 3.51 \text{s}\end{array}$