# Equation Demonstration -- Simple Pendulum Analysis

• Frankie
In summary: This requires a knowledge of the differential equation of simple harmonic motion and some calculus.In summary, the simple pendulum follows a harmonic oscillatory motion when the amplitude of the oscillations is small. This can be understood by looking at the differential equation for small oscillations and comparing it to the equation of motion for simple harmonic motion. Through some mathematical steps, it can be shown that the general solution for this differential equation is equivalent to the form given in image 2. This demonstration requires some knowledge of differential equations and calculus.
Frankie
Homework Statement
I need the demonstration of the simple pendulum "hourly law".
Relevant Equations
I've attached images
I've just studied simple pendulum: The simple pendulum (for small oscillations) differential equation is first image.
I've no problem to arrive this result and formula.
My problem is to get to the second formula by passing through another formula (Image 3) that my book mentions. I can't understand the passage.

Image 3 refers to acceleration expressed as a function of the position, is a simple harmonic motion.

In the image 3 last formula refers to centre where x0 = 0 and v0=ωA

My book: "For small oscillations the diffential equation becomes that of image 1 and coincides with that of the simple harmonic motion shown in image 3 set ω2 = g / L.
In conclusion, the motion of the pendulum is harmonic oscillatory when the amplitude of the oscillations is small so that senθ ≅ θ"
And then arrives at image2.

I would need to switch from the first to the second formula.

Image 1

Image 2

Image 3

Delta2
If ##\theta## is small, then the ##x##-coordinate of the pendulum bob is ##x = L\sin{\theta} \approx L \theta##, hence if ##\theta## satisfies ##\ddot{\theta} = -\omega^2 \theta##, then ##L\ddot{\theta} = -\omega^2 L\theta \implies \ddot{x} = -\omega^2 x## in this regime. The integral they wrote down gives$$\int_{x_0}^{x} \ddot{x} dx = \int_{x_0}^x -\omega^2 x dx = \frac{1}{2} \omega^2(x_0^2 - x^2)$$However, since ##dx = \dot{x} dt##, you could also write that very same integral as$$\int_{x_0}^{x} \ddot{x} dx = \int_{t_0}^{t} \ddot{x} \dot{x} dt = \int_{t_0}^t \frac{d}{dt} \left( \frac{1}{2} \dot{x}^2 \right) dt = \frac{1}{2}v^2 - \frac{1}{2}v_0^2$$Hence, equating these gives$$\frac{1}{2} \omega^2(x_0^2 - x^2) = \frac{1}{2}v^2 - \frac{1}{2}v_0^2 \implies v^2 = v_0^2 + \omega^2 (x_0^2 - x^2)$$Does that answer your questions?

Frankie
etotheipi said:
If ##\theta## is small, then the ##x##-coordinate of the pendulum bob is ##x = L\sin{\theta} \approx L \theta##, hence if ##\theta## satisfies ##\ddot{\theta} = -\omega^2 \theta##, then ##L\ddot{\theta} = -\omega^2 L\theta \implies \ddot{x} = -\omega^2 x## in this regime. The integral they wrote down gives$$\int_{x_0}^{x} \ddot{x} dx = \int_{x_0}^x -\omega^2 x dx = \frac{1}{2} \omega^2(x_0^2 - x^2)$$However, since ##dx = \dot{x} dt##, you could also write that very same integral as$$\int_{x_0}^{x} \ddot{x} dx = \int_{t_0}^{t} \ddot{x} \dot{x} dt = \int_{t_0}^t \frac{d}{dt} \left( \frac{1}{2} \dot{x}^2 \right) dt = \frac{1}{2}v^2 - \frac{1}{2}v_0^2$$Hence, equating these gives$$\frac{1}{2} \omega^2(x_0^2 - x^2) = \frac{1}{2}v^2 - \frac{1}{2}v_0^2 \implies v^2 = v_0^2 + \omega^2 (x_0^2 - x^2)$$Does that answer your questions?
At first thank you. You have already helped me a lot and the coincidence you demonstrated is clear, but now

i can't understand this:

on a theoretical-physical level I've understood it, I miss the mathematical step

how can i pass from $$\ddot{\theta} + \frac{g}{L} \theta = 0$$ to $$\theta = {\theta_0} \sin({ωt} +∅)$$

with ω2 = g / L.

Thanks again.

This is a second order homogenous differential equation in ##\theta(t)##. As with any differential equation of this form, you can trial solutions of the form ##e^{\lambda t}## and then write the complementary equation,$$\lambda^2 + \frac{g}{L} = 0 \implies \theta = \pm \sqrt{-\frac{g}{L}} = \pm i \sqrt{\frac{g}{L}} = \pm i \omega$$where we defined ##\omega := \sqrt{\frac{g}{L}}##. This means that our general solution is $$\theta(t) = Ae^{i \omega t} + Be^{-i \omega t} = (A+B)\cos{\omega t} + (A-B)i \sin{\omega t}$$or more simply, by collecting those coefficients into two arbitrary constants$$\theta(t) = C\cos{\omega t} + D\sin{\omega t} = E \sin{(\omega t + \phi)}$$Now you can use the boundary conditions to determine ##E## and ##\phi##.

Frankie
The pass is briefly as follows (this is in homework section so i cannot say the full details)

First we make two guesses that the functions ##f_1=\sin\omega t## and ##f_2=\cos\omega t## (##\omega^2=\frac{g}{L}##) verify the differential equation. Hence we know two solutions.

We know that these two solutions are linearly independent.

Then there is a theorem from the theory of vector spaces and differential equations that if for a linear differential equation of order ##n## we know ##n## linearly independent solutions ##f_1,f_2,...,f_n## then the general solution is $$f=\lambda_1f_1+\lambda_2f_2+...+\lambda_nf_n$$ where the ##\lambda_i## are any real numbers.

We apply this theorem for n=2 (the order of differential equation in this problem is 2). Then it is a matter of some trigonometry and simple algebra to prove that the general solution is equivalent to the form given in your post. (I see now that @etotheipi has posted something relevant to this).

Frankie and etotheipi
Frankie said:
Homework Statement:: I need the demonstration of the simple pendulum "hourly law".
Relevant Equations:: I've attached images

I would need to switch from the first to the second formula.

Image 1

Image 2

Just to simplify things a little. It is difficult to ab initio generate the solution to image 1 and obtain image 2...this is the stuff of entire courses on differential equations.
Happily for the purposes of this exercise it is necessary only to show that image 2 is a solution to image 1. This requires only the ability to take derivatives of trig functions. The rest is all correct but not really necessary and probably off-putting.

## 1. What is a simple pendulum?

A simple pendulum is a weight suspended from a fixed point that is able to swing back and forth due to the force of gravity. It consists of a mass (the weight) and a string or rod (the suspension) and is used to demonstrate the principles of oscillation and gravity.

## 2. How is the period of a simple pendulum calculated?

The period of a simple pendulum is calculated using the equation T = 2π√(L/g), where T is the period in seconds, L is the length of the pendulum in meters, and g is the acceleration due to gravity (9.8 m/s² on Earth). This equation assumes small angles of oscillation (less than 15°).

## 3. What factors affect the period of a simple pendulum?

The period of a simple pendulum is affected by the length of the pendulum, the mass of the weight, and the strength of gravity. The longer the pendulum, the longer the period. A heavier weight will also result in a longer period. The strength of gravity affects the period because it determines the acceleration of the pendulum.

## 4. How does the amplitude of a simple pendulum affect its period?

The amplitude (maximum angle of swing) of a simple pendulum does not affect its period as long as the angles are small (less than 15°). The period is only affected by the length of the pendulum, mass of the weight, and strength of gravity.

## 5. What is the purpose of demonstrating a simple pendulum?

The purpose of demonstrating a simple pendulum is to showcase the principles of oscillation and gravity. It can also be used to measure the strength of gravity in a particular location, as well as to study the relationship between the length and period of a pendulum. Simple pendulums are also commonly used in timekeeping devices such as grandfather clocks.

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