Simple Pendulum

Figure 4.5: Simple pendulum.

• The motion of a simple pendulum, which consists of a point mass $m$ suspended on the end of a rope of length $L$.
• Newton's second law for motion in the tangential direction:

  $$F=ma \longrightarrow -mgSin \theta=m \frac{d^2s}{dt^2}$$

• Here s is the arc length and $\theta$ is the angle the rope makes with the vertical (see Figure).

• Since $s=L\theta$,

  $$\frac{d^2\theta}{dt^2}=-\frac{g}{L}sin \theta$$

• This differential equation has no analytical solution.

• However, for small angle oscillations $sin \theta \approx \theta$ is approximated:

  $$\frac{d^2\theta}{dt^2}=-\frac{g}{L}\theta$$ (4.4)

• This equation has a solution in terms of sinusoidal functions:

  $$\theta(t)=\theta_0 cos\frac{2\pi t}{T}$$

  $$\boxed{T=2\pi \sqrt{\frac{L}{g}}}$$

• Here T is the period of the oscillation and $\theta_0$ is the angular amplitude.
• This formula is small angle ( $\theta_0 \le 15^\circ$) amplitudes gives approximately good results.
• We want to find the exact solution of the pendulum problem for each amplitude $\theta_0$ numerically.
• For this purpose, we will transform the problem into a numerical integral.

• Let's multiply both sides of the equation 4.4 by $d\theta /dt$ and rearrange:
  

  $$\frac{d\theta}{dt}\frac{d^2\theta}{dt^2} =-\frac{g}{L}sin \theta \frac{d\theta}{dt}$$

  $$\frac{1}{2}\frac{d}{dt}\left[ \frac{d\theta}{dt} \right]^2 =-\frac{g}{L}\frac{d} {dt} cos \theta$$

  
  
• If the indefinite integral is taken side by side,
  $$\frac{1}{2}\left( \frac{d\theta}{dt} \right)^2 =\frac{g}{L} cos \theta + C$$

• The initial velocity and angle values are used to determine the integration constant C.
  • If the pendulum is initially released from the maximum angle, its velocity will be zero.
  • So, if $d\theta /dt=0$ and $\theta=\theta_0$ are substituted into the equation at time $t=0$, $C= -(g/L) cos\theta_0$ is found.

• If the constant C is put in place and the arrangement is made,
  

  $$\frac{d\theta}{dt} =\sqrt{\frac{2g}{L}(cos\theta -cos\theta_0)}$$

  $$dt =\sqrt{\frac{L}{2g}} \frac{d\theta}{\sqrt{cos\theta -cos\theta_0}}$$

  
  
• Integrating the right side of this expression from $\theta=0$ to the angle $\theta=\theta_0$, the left side will be one quarter of a period.
  $$\frac{T}{4}= \sqrt{\frac{L}{2g}} \int_0^{\theta_0} \frac{d\theta}{\sqrt{cos\theta -cos\theta_0}}$$
  Thus, we have written the period formula as an integral.
• For each given amplitude $\theta_0$ we can calculate this integral numerically.
• However, since the value of the integrand diverges at the upper bound (for $\theta=\theta_0$), it is necessary to do a variable replacement first.

• For this purpose, first, the identities $cos\theta=1-2sin^2(\theta/2)$ and $cos\theta_0=1-2sin^2(\theta_0/2)$ are inserted in the denominator,
• and then with the variable change $k=sin (\theta_0/2)=\frac{sin (\theta/2)}{sin \phi}$, the integral becomes:

  $$\boxed{T= 4\sqrt{\frac{L}{g}} \int_0^{\pi/2} \frac{d\phi}{\sqrt{1-sin^2(\theta_0/2)sin^2\phi}}}$$

• This integral whose denominator is never zero is known as an elliptical integral of the first kind.
• There is no analytical solution.