“Tick-tock, tick-tock, tick-tock, tick-tock—” When we see the pendulum clock in our home swinging back and forth and reporting the time accurately, have we ever wondered about the mystery behind it?
One day, you want to play a prank on your mother by tying a weight to the pendulum, thinking that the clock will definitely run faster and throw her into confusion. However, you will soon find to your disappointment that the pendulum clock still keeps time accurately without any abnormality, as if time is declaring its uncontrollability. You feel very puzzled: why did my plan fail?
It is said that the first person in the world to study the simple pendulum was Galileo. Through multiple experiments, he concluded that the period of a simple pendulum depends only on the length of the string and is independent of the weight of the pendulum. Now you understand that to prank your mother, you should have increased the length of the pendulum... ^_^
Now let us analyze this simple pendulum...
Let the length of the simple pendulum be l. During the downward swing, considering only gravity, mechanical energy is conserved. Point A is the starting point, where the velocity is 0. When the pendulum falls to point B, the velocity is v, and the kinetic energy is 1/2 mv^2. The kinetic energy is converted from gravitational potential energy. Let RQ = h = l(\cos\theta - \cos\theta_0). The decrease in gravitational potential energy is mgh = mgl(\cos\theta - \cos\theta_0). Thus, we can list: \begin{aligned} v^2 &= 2gl(\cos\theta - \cos\theta_0) = \left(\frac{ds}{dt}\right)^2 = \left(l\frac{d\theta}{dt}\right)^2 \\ \frac{d\theta}{dt} &= \sqrt{\frac{2g}{l}}\sqrt{\cos\theta - \cos\theta_0} \end{aligned}
Using \cos\theta = 1 - 2\sin^2 \frac{\theta}{2}, it can be transformed into: \frac{dt}{d\theta} = \frac{1}{2} \sqrt{\frac{l}{g}} \frac{1}{\sqrt{\sin^2 \frac{\theta_0}{2} - \sin^2 \frac{\theta}{2}}}
For a pendulum, a period T is defined as the time taken to go from point A to point C and back to point A. Since the time taken from A to P is the same as from P to C, T = 4T_{AP}, i.e., T = 4 \int_0^{\theta_0} \frac{1}{2} \sqrt{\frac{l}{g}} \frac{d\theta}{\sqrt{\sin^2 \frac{\theta_0}{2} - \sin^2 \frac{\theta}{2}}}
In fact, for different \theta_0, the result of this integral is different. But why do high school textbooks say that the period is independent of the angle? When \theta is small, we use the relationship \sin^2 \frac{\theta}{2} \approx \left(\frac{\theta}{2}\right)^2. After substitution, it becomes: T = 4 \sqrt{\frac{l}{g}} \int_0^{\theta_0} \frac{d\theta}{\sqrt{\theta_0^2 - \theta^2}}
From the knowledge of definite integrals, we know the result is T = 2\pi \sqrt{\frac{l}{g}}, which is exactly Galileo’s result, independent of mass and the initial angle.
If a precise calculation is needed, one can think from the following direction:
Let \sin\frac{\theta_0}{2} = k, \sin\frac{\theta}{2} = k \sin\varphi, where \varphi \in (0, \frac{\pi}{2}). Then \frac{1}{2} \cos\frac{\theta}{2} d\theta = k \cos\varphi d\varphi, i.e., d\theta = \frac{2k \cos\varphi d\varphi}{\sqrt{1 - k^2 \sin^2 \varphi}}
After substitution, we get: T = 4 \sqrt{\frac{l}{g}} \int_0^{\frac{\pi}{2}} \frac{d\varphi}{\sqrt{1 - k^2 \sin^2 \varphi}}
This type of integral is called a “Complete Elliptic Integral of the First Kind”. Its result is: T = 2\pi \sqrt{\frac{l}{g}} \left(1 + \frac{1^2}{2^2} \sin^2 \frac{\theta_0}{2} + \frac{1^2 \cdot 3^2}{2^2 \cdot 4^2} \sin^4 \frac{\theta_0}{2} + \dots\right)
If only the first term is taken, it is the approximation formula from the beginning. The derivation process can be found in university “Mathematical Analysis” tutorials, or refer to the image below.
Mathematical Media - Detailed Analysis of the Simple Pendulum.pdf
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