Question

Suppose a pendulum is formed by attaching a massto theend of a string of negligible mass and length l. At$\mathit{t}\mathbf{=}\mathbf{0}$thependulum is released from rest at a small displacement angle${\mathit{\theta }}_{\mathbf{0}}\mathbf{>}\mathbf{0}$to the right of the vertical equilibrium position OP. SeeFigure 5.R.5. At time${\mathit{t}}_{\mathbf{1}}\mathbf{>}\mathbf{0}$the string hits a nail at a point N onOP a distance$\frac{\mathbf{3}}{\mathbf{4}}\mathit{l}$from O, but the mass continues to the left asshown in the figure.

(a) Construct and solve a linear initial-value problem for thedisplacement angleshown in the figure. Find theinterval$\mathbf{[}\mathbf{0}\mathbf{,}{\mathit{t}}_{\mathbf{1}}\mathbf{]}$on which${\mathit{\theta }}_{\mathbf{1}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$is defined.

(b) Construct and solve a linear initial-value problem for thedisplacement angle ${\mathit{\theta }}_{\mathbf{2}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$shown in the figure. Find theinterval$\mathbf{[}{\mathit{t}}_{\mathbf{1}}\mathbf{,}{\mathit{t}}_{\mathbf{2}}\mathbf{]}$on which ${\mathit{\theta }}_{\mathbf{2}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$is defined, where $t_2$isthe time that m returns to the vertical line NP.

Short answer

(a) The equation is $\frac{{\text{d}}^{\text{2}}{\text{θ}}_{\text{1}}}{{\text{dt}}^{\text{2}}}\text{+}\frac{\text{g}}{\text{l}}{\text{θ}}_{\text{1}}{\text{=0,θ}}_{\text{1}}{\text{(0)=θ}}_{\text{0}}{\text{,θ}}_1^{\text{'}}\text{(0)=0}$, its solution is ${\text{θ}}_{\text{1}}{\text{=θ}}_{\text{0}}\text{cos}\left(\sqrt{\frac{\text{g}}{\text{l}}}\text{t}\right)$, and the interval defined is .$\left[\text{0,}\sqrt{\frac{\text{l}}{\text{g}}}\frac{\text{π}}{\text{2}}\right]$

(b) The equation is $\frac{{\text{d}}^{\text{2}}{\text{θ}}_{\text{2}}}{{\text{dt}}^{\text{2}}}\text{+}\frac{\text{4g}}{\text{l}}{\text{θ}}_{\text{2}}{\text{=0,θ}}_{\text{2}}{\text{(0)=0,θ}}_2^{\text{'}}{\text{(0)=-θ}}_{\text{0}}\sqrt{\frac{\text{g}}{\text{l}}}$, its solution is${\text{θ}}_{\text{2}}\text{(t)=-}\frac{{\text{θ}}_{\text{0}}}{\text{2}}\text{sin}\left(\text{2}\sqrt{\frac{\text{g}}{\text{l}}}\text{t}\right)$ , and the interval defined is .$\left[\sqrt{\frac{{\displaystyle \text{l}}}{{\displaystyle \text{g}}}}\frac{{\displaystyle \text{π}}}{{\displaystyle \text{2}}}\text{,}\sqrt{\frac{{\displaystyle \text{l}}}{{\displaystyle \text{g}}}}\text{π}\right]$

Step-by-step solution

Solve a linear initial value problem.

The linearized pendulum equation is given by,

$\frac{{\text{d}}^{\text{2}}\text{θ}}{{\text{dt}}^{\text{2}}}\text{+}\frac{\text{g}}{\text{l}}\text{θ=0}$… (1)

(a)

As the pendulum is released from rest, so${\text{θ}}_1^{\text{'}}\text{(0)=0}$. And as the pendulum is released at an angle${\text{θ}}_{\text{0}}$, so${\text{θ}}_{\text{1}}{\text{(0)=θ}}_{\text{0}}$ Substitute the value in the equation (1).

$\frac{{\text{d}}^{\text{2}}{\text{θ}}_{\text{1}}}{{\text{dt}}^{\text{2}}}\text{+}\frac{\text{g}}{\text{l}}{\text{θ}}_{\text{1}}{\text{=0,θ}}_{\text{1}}{\text{(0)=θ}}_{\text{0}}{\text{,θ}}_1^{\text{'}}\text{(0)=0}$

Let the complementary equation be${\text{m}}^{\text{2}}\text{+}\frac{\text{g}}{\text{l}}\text{=0}$, and its roots be${\text{m}}_{\text{1,2}}\text{=±i}\sqrt{\frac{\text{g}}{\text{l}}}$. Then, the general solution for the equation is given by,

$\begin{array}{c}{\text{θ}}_{\text{1}}{\text{(t)=c}}_{\text{1}}\text{cos}\left(\sqrt{\frac{\text{g}}{\text{l}}}\text{t}\right){\text{+c}}_{\text{2}}\text{sin}\left(\sqrt{\frac{\text{g}}{\text{l}}}\text{t}\right)\\ {\text{θ}}_1^{\text{'}}{\text{(t)=-c}}_{\text{1}}\sqrt{\frac{\text{g}}{\text{l}}}\text{sin}\left(\sqrt{\frac{\text{g}}{\text{l}}}\text{t}\right){\text{+c}}_{\text{2}}\sqrt{\frac{\text{g}}{\text{l}}}\text{cos}\left(\sqrt{\frac{\text{g}}{\text{l}}}\right)\end{array}$

Apply the initial condition${\theta }_0={\theta }_1\left(0\right)=c_1\Rightarrow c_1={\theta }_0$and$0={\theta }_1^{\text{'}}\left(0\right)=c_2\sqrt{\frac{g}{l}}\Rightarrow c_2=0$in the above equation.

${\theta }_1\left(t\right)={\theta }_0\cos \left(\sqrt{\frac{g}{l}}t\right)$

Solve the interval.

Thus,${\theta }_1$is defined as long as it does not reach the equilibrium position. That is,${\theta }_1\left(t\right)$will be defined on$[0,t_1]$where $t_1$is the first solution to${\theta }_1\left(t\right)=0$

$\begin{array}{c}{\theta }_1\left(t\right)=0\\ {\theta }_0\cos \left(\sqrt{\frac{g}{l}}t\right)=0\\ \sqrt{\frac{g}{l}}t=\frac{\pi }{2}\\ t_1=\sqrt{\frac{l}{g}}\frac{\pi }{2}{\theta }_1\left(t\right)\end{array}$

Hence, ${\theta }_1\left(t\right)$will be defined on the interval .$\left[\text{0,}\sqrt{\frac{{\displaystyle \text{l}}}{{\displaystyle \text{g}}}}\frac{{\displaystyle \text{π}}}{{\displaystyle \text{2}}}\right]$

Solve a linear initial value problem.

Once the pendulum reaches the nail, the new displacement is${\theta }_2$corresponds to the pendulum with a different length which is equal to$\frac{l}{4}$. When this change happens, the angle is equal to zero, and the velocity is equal to,

${\theta }_1^{\text{'}}\left(t_1\right)=-{\theta }_0\sqrt{\frac{g}{l}}\sin \left(\sqrt{\frac{{\displaystyle g}}{{\displaystyle l}}}\sqrt{\frac{{\displaystyle l}}{{\displaystyle g}}}\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)=-\sqrt{\frac{g}{l}}{\theta }_0\sin \left(\frac{\pi }{2}\right)=-{\theta }_0\sqrt{\frac{g}{l}}$

The initial value problem is given by,

$\begin{array}{l}\frac{{\text{d}}^{\text{2}}{\text{θ}}_{\text{2}}}{{\text{dt}}^{\text{2}}}\text{+}\frac{\text{g}}{\frac{\text{l}}{\text{4}}}{\text{θ}}_{\text{2}}{\text{=0,θ}}_{\text{2}}{\text{(0)=0,θ}}_2^{\text{'}}{\text{(0)=-θ}}_{\text{0}}\sqrt{\frac{\text{g}}{\text{l}}}\\ \frac{{\text{d}}^{\text{2}}{\text{θ}}_{\text{2}}}{{\text{dt}}^{\text{2}}}\text{+}\frac{\text{4g}}{\text{l}}{\text{θ}}_{\text{2}}{\text{=0,θ}}_{\text{2}}{\text{(0)=0,θ}}_2^{\text{'}}{\text{(0)=-θ}}_{\text{0}}\sqrt{\frac{\text{g}}{\text{l}}}\end{array}$

The general solution for the equation is given by,

$\begin{array}{l}{\text{θ}}_{\text{2}}{\text{(t)=c}}_{\text{1}}\text{cos}\left(\text{2}\sqrt{\frac{\text{g}}{\text{l}}}\text{t}\right){\text{+c}}_{\text{2}}\text{sin}\left(\text{2}\sqrt{\frac{\text{g}}{\text{l}}}\text{t}\right)\\ {\text{θ}}_2^{\text{'}}{\text{(t)=-2c}}_{\text{1}}\text{sin}\left(\text{2}\sqrt{\frac{\text{g}}{\text{l}}\text{t}}\right){\text{+2c}}_{\text{2}}\sqrt{\frac{\text{g}}{\text{l}}}\text{cos}\left(\text{2}\sqrt{\frac{\text{g}}{\text{l}}}\text{t}\right)\end{array}$

Apply the initial condition${\text{0=θ}}_{\text{2}}{\text{(0)=c}}_{\text{1}}\Rightarrow {\text{c}}_{\text{1}}\text{=0}$and${\text{-θ}}_{\text{0}}\sqrt{\frac{\text{g}}{\text{l}}}{\text{=θ}}_2^{\text{'}}{\text{(0)=2c}}_{\text{2}}\sqrt{\frac{\text{g}}{\text{l}}}\Rightarrow {\text{c}}_{\text{2}}\text{=-}\frac{\text{1}}{\text{2}}{\text{θ}}_{\text{0}}$in the above equation.

${\text{θ}}_{\text{2}}\text{(t)=-}\frac{{\text{θ}}_{\text{0}}}{\text{2}}\text{sin}\left(\text{2}\sqrt{\frac{\text{g}}{\text{l}}}\text{t}\right)$

Solve the interval.

Thus,${\theta }_2$is defined as long as the pendulum reaches the vertical line ${\theta }_2\left(t\right)=0$. That is, will be defined on $[t_1,t_2]$where$t_2$is the first solution to${\theta }_2\left(t\right)=0$.

$\begin{array}{c}{\theta }_2\left(t\right)=0\\ -\frac{{\theta }_0}{2}\sin \left(2\sqrt{\frac{g}{l}}t\right)=0\\ \sin \left(2\sqrt{\frac{g}{l}}t\right)=0\\ 2\sqrt{\frac{g}{l}}t=n\pi \\ t=\sqrt{\frac{l}{g}\frac{n\pi }{2}}\end{array}$

where is an positive integer. The first positive solution larger than$\sqrt{\frac{l}{g}}\frac{\pi }{2}$ occurs for.$n=2$ So,

$t_2=\sqrt{\frac{l}{g}\pi }$

Hence,${\theta }_2\left(t\right)$ will be defined on the interval role="math" localid="1664175666919" $\left[\sqrt{\frac{\text{l}}{\text{g}}}\frac{\text{π}}{\text{2}}\text{,}\sqrt{\frac{\text{l}}{\text{g}}}\text{π}\right]$.