Question

Spring Pendulum The rotational form of Newton’s secondlaw of motion is:The time rate of change of angular momentum about a point isequal to the moment of the resultant force (torque).In the absence of damping or other external forces, an analogueof (14) in Section 5.3 for the pendulum shown in Figure 5.3.3is then

(a) When m and l are constant show that (1) reduces to (6) ofSection 5.3.(b) Now suppose the rod in Figure 5.3.3 is replaced with aspring of negligible mass. When a mass m is attached toits free end the spring hangs in the vertical equilibriumposition shown in Figure 5.R.4 and has length l0. When the spring pendulum is set in motion we assume that themotion takes place in a vertical plane and the spring is stiffenough not to bend. For t . 0 the length of the spring isthen lstd 5 l0 1 xstd, whereis the displacement from theequilibrium position. Find the differential equation for thedisplacement angledefined by (1).

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 $\left[\text{0,}\sqrt{\frac{{\displaystyle \text{l}}}{{\displaystyle \text{g}}}}\frac{{\displaystyle \text{π}}}{{\displaystyle \text{2}}}\right]$.

(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,

role="math" localid="1664189128139" $\begin{array}{c}{\text{θ}}_{\text{1}}{\text{(t)=c}}_{\text{1}}\text{cos}\left(\sqrt{\frac{\text{g}}{\text{l}}}\text{t}\right){\text{+c}}_{\text{2}}\text{sin}\left(\sqrt{\frac{\text{g}}{\text{l}}}\text{t}\right)\\ {\text{θ}}_1^{\text{'}}{\text{(t)=-c}}_{\text{1}}\sqrt{\frac{\text{g}}{\text{l}}}\text{sin}\left(\sqrt{\frac{\text{g}}{\text{l}}}\text{t}\right){\text{+c}}_{\text{2}}\sqrt{\frac{\text{g}}{\text{l}}}\text{cos}\left(\sqrt{\frac{\text{g}}{\text{l}}}\right)\end{array}$

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.

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

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 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, will be defined on the interval role="math" localid="1664189434043" $\left[\text{0,}\sqrt{\frac{\text{l}}{\text{g}}}\frac{\text{π}}{\text{2}}\right]$.

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,

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

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, ${\theta }_2\left(t\right)$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 n 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$\left[\sqrt{\frac{{\displaystyle \text{l}}}{{\displaystyle \text{g}}}}\frac{{\displaystyle \text{π}}}{{\displaystyle \text{2}}}\text{,}\sqrt{\frac{{\displaystyle \text{l}}}{{\displaystyle \text{g}}}}\text{π}\right]$ .