Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 67

(a) What is the energy of a pendulum \((L=1.0 \mathrm{m}, m=\) $0.50 \mathrm{kg})\( oscillating with an amplitude of \)5.0 \mathrm{cm} ?$ (b) The pendulum's energy loss (due to damping) is replaced in a clock by allowing a \(2.0-\mathrm{kg}\) mass to drop \(1.0 \mathrm{m}\) in 1 week. What average percentage of the pendulum's energy is lost during one cycle?

Short Answer

Expert verified
Answer: The average percentage of energy loss during one cycle for a pendulum with a given mass and amplitude is given by the formula: Percent energy loss = \( \frac{256}{T} \% \) where T is the period of the pendulum's oscillation.

Step by step solution

01

Calculate the maximum potential energy of the pendulum

The formula for potential energy is: \(PE = mgh\) ,where m is the mass of the pendulum (0.5 kg), g is the acceleration due to gravity (9.81 \(m/s^2\)) and h is the maximum height of the pendulum during its oscillation. We can calculate h using amplitude: $$ h = \sqrt{L^2 - (L-A)^2} $$ where A is the amplitude (5.0 cm = 0.05 m) and L is the length of the pendulum (1.0 m). Calculating the maximum height, we have: $$ h = \sqrt{(1.0)^2 - (1.0-0.05)^2} = \sqrt{1-0.95^2} = 0.15605025 m $$ Now, calculate the maximum potential energy: $$ PE_{max} = mgh = (0.50 kg)(9.81 m/s^2)(0.15605025 m) ≈ 0.765 J $$
02

Calculate the energy loss during one cycle

The pendulum loses some of its energy due to damping and is replaced by allowing a 2.0 kg mass to drop 1.0 m during one week. We can calculate the energy replaced in one cycle. Let T be the period of the pendulum's oscillation, then the number of oscillation cycles per week is \(N = \frac{7 \times 24 \times 60 \times 60}{T}\). Now, we calculate the energy loss during one week: $$ E_{week} = mgh = (2.0 kg)(9.81 m/s^2)(1.0 m) = 19.62 J $$ Now we can find the energy loss during one cycle: $$ E_{cycle} = \frac{E_{week}}{N} = \frac{19.62 J}{\frac{7\times 24\times 60\times 60}{T}} = 19.62 J \times \frac{T}{7\times 24\times 60\times 60} $$
03

Calculate the average percentage of energy loss during one cycle

To find the percentage of energy loss during one cycle, we can use the following formula: Percent energy loss = \(\frac{E_{cycle}}{PE_{max}} \times 100\) Substituting the values, we have: $$ \text{Percent energy loss} = \frac{19.62 J \times \frac{T}{7\times 24\times 60\times 60}}{0.765 J} \times 100 = \frac{256}{T} \% $$ Since we don't know the period (T), we cannot provide an exact representation, but we provided the percentage of energy loss during one cycle for any given period T.

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

In an aviation test lab, pilots are subjected to vertical oscillations on a shaking rig to see how well they can recognize objects in times of severe airplane vibration. The frequency can be varied from 0.02 to $40.0 \mathrm{Hz}\( and the amplitude can be set as high as \)2 \mathrm{m}$ for low frequencies. What are the maximum velocity and acceleration to which the pilot is subjected if the frequency is set at \(25.0 \mathrm{Hz}\) and the amplitude at \(1.00 \mathrm{mm} ?\)
A \(0.50-\mathrm{kg}\) object, suspended from an ideal spring of spring constant \(25 \mathrm{N} / \mathrm{m},\) is oscillating vertically. How much change of kinetic energy occurs while the object moves from the equilibrium position to a point \(5.0 \mathrm{cm}\) lower?
The period of oscillation of a spring-and-mass system is \(0.50 \mathrm{s}\) and the amplitude is \(5.0 \mathrm{cm} .\) What is the magnitude of the acceleration at the point of maximum extension of the spring?
The upper surface of a cube of gelatin, \(5.0 \mathrm{cm}\) on a side, is displaced \(0.64 \mathrm{cm}\) by a tangential force. If the shear modulus of the gelatin is \(940 \mathrm{Pa},\) what is the magnitude of the tangential force?
The gravitational potential energy of a pendulum is \(U=m g y .\) (a) Taking \(y=0\) at the lowest point, show that \(y=L(1-\cos \theta),\) where \(\theta\) is the angle the string makes with the vertical. (b) If \(\theta\) is small, \((1-\cos \theta)=\frac{1}{2} \theta^{2}\) and \(\theta=x / L\) (Appendix A.7). Show that the potential energy can be written \(U=\frac{1}{2} k x^{2}\) and find the value of \(k\) (the equivalent of the spring constant for the pendulum).
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Electricity

Read Explanation

Turning Points in Physics

Read Explanation

Particle Model of Matter

Read Explanation

Energy Physics

Read Explanation

Rotational Dynamics

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