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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: $$k = mg\frac{1}{L}.$$

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