Question

Consider a pendulum of length L. Let x (t) be the angle of the pendulum makes with the vertical (measured in radians).For small angles, the motion is well approximated by the DE

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{t}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{-}\mathbf{g}}{\mathbf{L}}\mathit{x}$

where g is the acceleration due to gravity$\mathbf{(}\mathit{g}\mathbf{\approx }\mathbf{9}\mathbf{.}\mathbf{81}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}\mathbf{)}$How long does the pendulum have to be so that swings from one extreme position to the other in exactly one second?

Short answer

The pendulum have to be so that swings from one extreme position to the other in exactly one second is one

Step-by-step solution

Explanation for Linear Differential operators and linear differential equations

A transformation T from ${\mathrm{C}}^{\infty }\mathrm{to}{\mathrm{C}}^{\infty }$of the form $T\left(f\right)=f^{\left(n\right)}+a_{n-1}{f}^{(n-1)}+...+a_{1}f\text{'}+a_{0}f$is called an nth-order linear differential operator. Here $f^{\left(k\right)}$denote the k-th derivative of function f and the coefficients $a_k$are complex scalars.

If T is an nth-order linear differential operator and g is a smooth function, then the equation becomes

$T\left(f\right)=g$

(or)

$f^{\left(n\right)}+a_{n-1}{f}^{(n-1)}+...+a_{1}f\text{'}+a_{0}f=g$

Is called an nth-order linear differential equation (DE).The DE is called homogeneous if g = 0 and inhomogeneous otherwise.

Explanation for the solution of the pendulum problem

Consider a pendulum with varying length. Here the pendulum is formed by a mass m attached to the end of the wire which is attached to the ceiling.

Assume that the length l (t) of the wire varies with time.

Let the angle in radians be denoted as $\theta \left(t\right)$which is between the pendulum and the vertical, then the motion of the pendulum is governed for small angles by the initial value problem as

$$\begin{gathered} l^2\left(t\right)\theta \text{'}\text{'}\left(t\right)+2l\left(t\right)l\text{'}\left(t\right)+gl\left(t\right)\sin \left(\theta \right(\left(t\right))=0 \\ \theta (0)={\theta }_0,\theta \text{'}(0)={\theta }_1 \end{gathered}$$

Where g is the acceleration due to gravity.

Assume that

$l\left(t\right)=l_0+l_1\cos (\omega t-f)$

Where$l_1$is much smaller than$l_0$

And also the pendulum takes one second to swing from one extreme to the other ,If it has the ability to swing freely, then it will take another second to swing back to its starting position.

Therefore, a pendulum takes two seconds to complete one full cycle of movement or one oscillation,

Hence the solution.