Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 51

An \(80.0-\mathrm{kg}\) bungee jumper is enjoying an afternoon of jumps. The jumper's first oscillation has an amplitude of \(10.0 \mathrm{~m}\) and a period of \(5.00 \mathrm{~s}\). Treating the bungee cord as spring with no damping, calculate each of the following: a) the spring constant of the bungee cord. b) the bungee jumper's maximum speed during the ascillation, and c) the time for the amplitude to decrease to \(2.00 \mathrm{~m}\) (with air resistance providing the damping of the oscillations at \(7.50 \mathrm{~kg} / \mathrm{s})\)

Short Answer

Expert verified
Question: A bungee jumper with a mass of 80.0 kg experiences simple harmonic motion with a period of 5.00 s and an initial amplitude of 10.0 m. The damping rate due to air resistance is 7.50 kg/s. Calculate a) the spring constant of the bungee cord, b) the maximum speed of the jumper during oscillation, and c) the time taken for the amplitude to reduce to 2.00 m. Answer: a) The spring constant is approximately 200 N/m. b) The maximum speed of the jumper during oscillation is approximately 12.6 m/s. c) The time taken for the amplitude to reduce to 2.00 m is approximately 23.5 s.

Step by step solution

01

Part a: Finding the spring constant

Using the formula for the period of oscillation: \(T = 2\pi\sqrt{\frac{m}{k}}\), the spring constant (k) can be found by rearranging the equation and solving for k. Given that the period \(T = 5.00s\) and the mass \(m = 80.0kg\), we can plug these values into our equation and solve for k: \(k = \frac{4\pi^2 m}{T^2} = \frac{4\pi^2 (80.0)}{(5.00)^2}\) The value we get for the spring constant, k, is approximately 200 N/m.
02

Part b: Finding the maximum speed

To find the maximum speed during the oscillation, we first need to find the angular frequency (\(\omega = \frac{2\pi}{T}\)). Given that the period \(T = 5.00s\), we calculate the angular frequency as: \(\omega = \frac{2\pi}{5.00s} \approx 1.26 s^{-1}\) Now, using the formula for the maximum speed (\(v_{max} = A\omega\)) and given that the amplitude \(A = 10.0m\), we can calculate the maximum speed as: \(v_{max} = (10.0m)(1.26 s^{-1}) \approx 12.6 m/s\)
03

Part c: Time for amplitude to decrease to 2.00m

Given that air resistance provides damping with the rate 7.50 kg/s, the amplitude's decay with time can be described by the following equation: \(A(t) = A_0 e^{-\frac{b}{2m}t}\) Here, \(A_0 = 10.0m\), \(A(t) = 2.00m\), \(b = 7.50kg/s\), and \(m = 80.0kg\). We need to solve for the time \(t\). Rearrange the equation and solve for time: \(t = \frac{-2m\ln{\frac{A(t)}{A_0}}}{b} = \frac{-2(80.0)\ln{\frac{2.00}{10.0}}}{7.50}\) The time it takes for the amplitude to decrease to 2.00m is approximately 23.5s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A small mass, \(m=50.0 \mathrm{~g}\), is attached to the end of a massless rod that is hanging from the ceiling and is free to swing. The rod has length \(L=1.00 \mathrm{~m} .\) The rod is displaced \(10.0^{\circ}\) from the vertical and released at time \(t=0\). Neglect air resistance. What is the period of the rod's oscillation? Now suppose the entire system is immersed in a fluid with a small damping constant, \(b=0.0100 \mathrm{~kg} / \mathrm{s},\) and the rod is again released from an initial displacement angle of \(10.0^{\circ}\). What is the time for the amplitude of the oscillation to reduce to \(5.00^{\circ}\) ? Assume that the damping is small Also note that since the amplitude of the oscillation is small and all the mass of the pendulum is at the end of the rod, the motion of the mass can be treated as strictly linear, and you can use the substitution \(R \theta(t)=x(t),\) where \(R=1.0 \mathrm{~m}\) is the length of the pendulum rod.

A mass of \(10.0 \mathrm{~kg}\) is hanging by a steel wire \(1.00 \mathrm{~m}\) long and \(1.00 \mathrm{~mm}\) in diameter. If the mass is pulled down slightly and released, what will be the frequency of the resulting oscillations? Young's modulus for steel is \(2.0 \cdot 10^{11} \mathrm{~N} / \mathrm{m}^{2}\)

The period of oscillation of an object in a frictionless tunnel running through the center of the Moon is \(T=2 \pi / \omega_{0}\) \(=6485 \mathrm{~s}\), as shown in Fxample 142 . What is the period of oscillation of an object in a similar tunnel through the Earth \(\left(R_{\mathrm{I}}=6.37 \cdot 10^{6} \mathrm{~m} ; R_{\mathrm{M}}=1.74 \cdot 10^{6} \mathrm{~m} ; M_{\mathrm{E}}=5.98 \cdot 10^{24} \mathrm{~kg}\right.\) \(\left.M_{u}=7.36 \cdot 10^{22} \mathbf{k g}\right) ?\)

A mass \(M=0.460 \mathrm{~kg}\) moves with an initial speed \(v=3.20 \mathrm{~m} / \mathrm{s}\) on a level frictionless air track. The mass is initially a distance \(D=0.250 \mathrm{~m}\) away from a spring with \(k=\) \(840 \mathrm{~N} / \mathrm{m}\), which is mounted rigidly at one end of the air track. The mass compresses the spring a maximum distance \(d\), before reversing direction. After bouncing off the spring the mass travels with the same speed \(v\), but in the opposite dircction. a) Determine the maximum distance that the spring is compressed. b) Find the total elapsed time until the mass returns to its starting point. (Hint: The mass undergoes a partial cycle of simple harmonic motion while in contact with the spring.)

When the amplitude of the oscillation of a mass on a stretched string is increased, why doesn't the period of oscil lation also increase?
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Medical Physics

Read Explanation

Magnetism

Read Explanation

Work Energy and Power

Read Explanation

Geometrical and Physical Optics

Read Explanation

Space Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free