Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 94

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).

Short Answer

Expert verified
Answer: The height (y) of the pendulum above its lowest point is given by the formula \(y = L(1 - \cos{\theta})\), where L is the length of the pendulum string, and \(\theta\) is the angle it makes with the vertical axis. The equivalent spring constant (k) for the pendulum is given by the formula \(k = mg\frac{1}{L}\), where m is the mass of the pendulum and g is the acceleration due to gravity.

Step by step solution

01

Draw and Analyze the Pendulum Geometry

Let's first draw the pendulum and mark important points and angles. Consider the pendulum in its initial position when the string makes an angle \(\theta\) with the vertical axis. Let O be the point from which the string is hanging, and A and B be the points on the string at angle \(\theta\) and \(90^\circ - \theta\) w.r.t. Point O respectively. Now we have a right-angled triangle OAB, with angle \(\theta\) at O, angle \(90^\circ - \theta\) at A, and angle \(90^\circ\) at B proving that it is a right triangle.
02

Apply Trigonometry to Find the Height

In a right-angled triangle, we can use the cosine function to find the side adjacent (OA) to the given angle \(\theta\): $$\cos{\theta} = \frac{OA}{L},$$ where L is the total length of the pendulum string. Then, we find that $$OA = L\cos{\theta}.$$ To find the height (y) of the pendulum above the lowest point, we need to calculate the vertical distance OB, which can be represented as $$y = L - OA.$$ Substituting OA from the previous step, we get: $$y = L(1 - \cos{\theta}).$$ #Part (b)#
03

Apply the Given Approximations

For small angles, we can use the given approximation $$(1 - \cos\theta) \approx \frac{1}{2}\theta^2$$ and $$\theta \approx \frac{x}{L}.$$ Substituting the second approximation into the first one, we get: $$(1 - \cos\theta) \approx \frac{1}{2}\left(\frac{x}{L}\right)^2.$$
04

Plug the Approximations into the Potential Energy Formula

Now, we will use the given potential energy formula, $$U = mgy.$$ Substituting y and the approximation from the previous step, we get: $$U = mgL\left(\frac{1}{2}\left(\frac{x}{L}\right)^2\right) = \frac{1}{2}mg\frac{x^2}{L}.$$
05

Find the Equivalent Spring Constant

Comparing this potential energy expression to the spring potential energy formula, $$U = \frac{1}{2}kx^2,$$ we conclude that the equivalent spring constant \(k\) is given by: $$k = mg\frac{1}{L}.$$

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

Resilin is a rubber-like protein that helps insects to fly more efficiently. The resilin, attached from the wing to the body, is relaxed when the wing is down and is extended when the wing is up. As the wing is brought up, some elastic energy is stored in the resilin. The wing is then brought back down with little muscular energy, since the potential energy in the resilin is converted back into kinetic energy. Resilin has a Young's modulus of $1.7 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}$ (a) If an insect wing has resilin with a relaxed length of \(1.0 \mathrm{cm}\) and with a cross-sectional area of \(1.0 \mathrm{mm}^{2},\) how much force must the wings exert to extend the resilin to \(4.0 \mathrm{cm} ?\) (b) How much energy is stored in the resilin?
A pendulum is made from a uniform rod of mass \(m_{1}\) and a small block of mass \(m_{2}\) attached at the lower end. (a) If the length of the pendulum is \(L\) and the oscillations are small, find the period of the oscillations in terms of \(m_{1}, m_{2}, L,\) and \(g .\) (b) Check your answer to part (a) in the two special cases \(m_{1} \gg m_{2}\) and \(m_{1}<<m_{2}\).
The amplitude of oscillation of a pendulum decreases by a factor of 20.0 in \(120 \mathrm{s}\). By what factor has its energy decreased in that time?
Show, using dimensional analysis, that the frequency \(f\) at which a mass- spring system oscillates radical is independent of the amplitude \(A\) and proportional to Whim. IHint: Start by assuming that \(f\) does depend on \(A\) (to some power).]
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?
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Force

Read Explanation

Classical Mechanics

Read Explanation

Space Physics

Read Explanation

Scientific Method Physics

Read Explanation

Magnetism

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