Given the motion of a pendulum modeled by:

$x = c_1\left[ \begin{array}{cccc} \cos2t\\-2\sin2t\end{array} \right] + c_2\left[ \begin{array}{cccc} \sin2t\\2\cos2t \end{array} \right]$

What initial conditions will result in the motion of the pendulum having a maximum positive outward displacement of one radian at $t = \pi/2$?

What does it means for the pendulum to have a "maximum displacement of one radian?" If I plug in $t = \pi/2$ into the system I get:

$x(\pi/2) = c_1\left[ \begin{array}{cccc} -1\\0\end{array} \right] + c_2\left[ \begin{array}{cccc} 0\\-2 \end{array} \right]$

but I'm not sure what this does for me. Should I solve the system

$\left[ \begin{array}{cccc} 1\\1\end{array} \right] = c_1\left[ \begin{array}{cccc} -1\\0\end{array} \right] + c_2\left[ \begin{array}{cccc} 0\\-2 \end{array} \right]$

for $c_1$ and $c_2$?


  • $\begingroup$ Displacement angle $=\sin^{-1}(\sqrt(x^2+y^2)/l$ where $l=$length of string. "maximum displacement of one radian" means that the displacement angle is at its greatest at $t=\frac{pi}2$ and is one radian. $\endgroup$ Nov 12, 2012 at 11:12

1 Answer 1


I guess that you mean the following: The maximal displacement is one (not "one radian"), and it should happen (in particular) for $t={\pi\over2}$.

Put $(c_1,c_2)=r(\cos\gamma,\sin\gamma)$ with $r\geq0$. Then $$\eqalign{x(t)&=\bigl(c_1\cos(2t)+c_2\sin(2t), -2c_1\sin(2t)+2c_2\cos(2t)\bigr) \cr &=r\bigl(\cos(\gamma-2t),2\sin(\gamma-2t)\bigr)\ .\cr}$$ Introduce the auxiliary variable $\tau:= \gamma-2t$. Then we have $$|x(t)|^2=r^2(\cos^2\tau +4\sin^2\tau)=r^2(1+3\sin^2\tau)\ .$$ It follows that $|x(t)|$ is maximal when $|\sin\tau|=1$ and has maximal value $2r$. Since this maximal value should be $=1$ we see that necessarily $r={1\over2}$.

Since we should have a maximal displacement at $t={\pi\over2}$ we want $$1=|\sin\tau|=\bigl|\sin(\gamma-2{\pi\over2})\bigl|=|\sin\gamma|\ ,$$ and this implies $\gamma=\pm{\pi\over2}$. This gives $(c_1,c_2)=\bigl(0,\pm{1\over2}\bigr)$. As $x(0)=(c_1,2c_2)$ the initial conditions that guarantee the required behavior are $x(0)=(0,\pm1)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .