Pendulum with Varying Length. A pendulum is formed by a mass m attached to the end of a wire that is attached to the ceiling. Assume that the length $I(t)$ of the wire varies with time in some predetermined fashion. If $\theta(t)$ is the angle in radians between the pendulum and the vertical, then the motion of the pendulum is governed for small angles by the initial value problem

$$\begin{array}{l}{l^{2}(t) \theta^{\prime \prime}(t)+2 l(t) l^{\prime}(t) \theta^{\prime}(t)+g l(t) \sin (\theta(t))=0} \\ {\theta(0)=\theta_{0}, \quad \theta^{\prime}(0)=\theta_{1}}\end{array}$$

where g is the acceleration due to gravity. Assume that

$$l(t)=I_{0}+l_{1} \cos (\omega t-\phi)$$

where $l_{1}$ is much smaller than $l_{0}$ l0. (This might be a model

for a person on a swing, where the $pumping$ action

changes the distance from the center of mass of the swing to the point where the swing is attached.) To simplify the computations, take $g=1$Using the Runge-Kutta algorithm with $h=0.1$ study the motion of the pendulum when $\theta_{0}=0.05, \theta_{1}=0, \quad l_{0}=1, l_{1}=0.1$ $\omega=1,$ and $\phi=0.02 .$ In particular, does the pendulum

ever attain an angle greater in absolute value than the

initial angle $\theta_{0} ?$

## Discussion

## Video Transcript

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## Recommended Questions

Period of a Pendulum The period $T($ in seconds) of a simple pendulum is a function of its length $l$ (in feet) defined by the equation

$$

T=2 \pi \sqrt{\frac{l}{g}}

$$

$$

\begin{array}{l}{\text { where } g \approx 32.2 \text { feet per second per second is the acceleration }} \\ { \text { due to gravity. (Continued on page }260 .)}\end{array}

$$

$$

\begin{array}{l}{\text { (a) Use a graphing utility to graph the function } T=T(l)} \\ {\text { (b) Now graph the functions } T=T(l+1), T=T(l+2)} \\ {\text { and } T=T(l+3)} \\ {\text { (c) Discuss how adding to the length } l \text { changes the period } T} \\ {\text { (d) Now graph the functions } T=T(2 l), T=T(3 l), \text { and }} \\ {T=T(4 l) .} \\ {\text { (e) Discuss how multiplying the length } l \text { by factors of } 2,3,} \\ {\text { and } 4 \text { changes the period } T}\end{array}

$$

In Section 5 of "oscillations," the oscillation of a simple pendulum (Fig. 46$)$ is viewed as linear motion along the arc length $x$ and analyzed via $F=m a$ . Alternatively, the pendulum's movement can be regarded as rotational motion about its point of support and analyzed using $\tau=I \alpha$ . Carry out this alternative analysis and show that

$\theta(t)=\theta_{\max } \cos \left(\sqrt{\frac{g}{\ell}} t+\phi\right)$

where $\quad \theta(t)$ is the angular displacement of the pendulum from the vertical at time $t,$ as long as its maximum value is less than about $15^{\circ}$ .

The period of a simple pendulum, defined as the time necessary for one complete oscillation, is measured in time units and is given by

$$

T=2 \pi \sqrt{\frac{\ell}{g}}

$$

where $\ell$ is the length of the pendulum and $g$ is the acceleration due to gravity, in units of length divided by time squared. Show that this equation is dimensionally consistent. (You might want to check the formula using your keys at the end of a string and a stopwatch.)

(II) A physical pendulum consists of an 85 -cm-long, 240 -g-mass, uniform wooden rod hung from a nail near one end (Fig, 38 ). The motion is damped because of friction in the pivot; the damping force is approximately proportional to $d \theta / d t .$ The rod is set in oscillation by displacing it $15^{\circ}$ from its equilibrium position and releasing it. After 8.0 $\mathrm{s}$ the amplitude of the oscillation has been reduced to $5.5^{\circ} .$ If the angular displacement can be written as $\theta=A e^{-\gamma t} \cos \omega^{\prime} t,$ find $(a) \gamma,(b)$ the approximate period of the motion, and $(c)$ how long it takes for the amplitude to be reduced to $\frac{1}{2}$ of its original value.

Use the result of Problem 8 to prove that if the pendulum in Figure 4.18 on page 208 is released from rest at the angle $ \alpha, 0<\alpha<\pi $ then $ |\theta(t)| \leq \alpha $ for all $ t $ [Hint: The initial conditions are $ \theta(0)=\alpha, \theta^{\prime}(0)=0 $; argue that the constant in Problem 8 equals $ -(g / \ell) \cos \alpha $ .]

The figure shows a pendulum with length $ L $ and the angle $ \theta $ from the vertical to the pendulum. It can be shown that $ \theta $, as a function of time, satisfies the nonlinear differential equation $$ \dfrac{d^2\theta}{dt^2} + \dfrac{g}{L} \sin \theta = 0 $$

where $ g $ is the acceleration due to gravity. For small values of $ \theta $ we can use the linear approximation $ \sin \theta \approx \theta $ and then the differential equation becomes linear.

(a) Find the equation of motion of a pendulum with length 1 m if $ \theta $ is initially 0.2 rad and the initial angular velocity is $ d\theta/dt = 1 $ rad/s.

(b) What is the maximum angle from the vertical?

(c) What is the period of the pendulum (that is, the time to complete one back-and-forth swing)?

(d) When will the pendulum first be vertical?

(e) What is the angular velocity when the pendulum is vertical?

On page 431 of Physics: Calculus, 2 $\mathrm{d}$ ed. by Eugene Hecht

(Pacific Grove, $\mathrm{CA}, 2000 ),$ in the course of deriving the formula $T=2 \pi \sqrt{L / g}$ for the period of a pendulum of length

$L,$ the author obtains the equation $a_{T}=-g \sin \theta$ for the tangential acceleration of the bob of the pendulum. He then says, "for small angles, the value of $\theta$ in radians is very

nearly the value of $\sin \theta ;$ they differ by less than 2$\%$ out to

about $20^{\circ} . "$

(a) Verify the linear approximation at 0 for the sine function:

$\sin x \approx x$

(b) Use a graphing device to determine the values of $x$ for

which sin $x$ and $x$ differ by less than 2$\%$ . Then verify

Hecht's statement by converting from radians to degrees.

On page 431 of $ Physics: Calculus, $ 2d ed., by Eugene Hecht (Pacific Grove, CA: Brooks/ Cole, 2000), in the course of deriving the formula $ T = 2\pi \sqrt {L/g} $ for the period of a pendulum of length $ L, $ the author obtains the equation $ a_T = -g \sin \theta $ for the tangential acceleration of the bob of the pendulum. He then says, "for small angles, the value of $ \theta $ in radians is very nearly the value of $ \sin \theta; $ they differ by less than $ 2\% $ out to about $ 20^o." $

(a) Verify the linear approximation at 0 for the sine function:

$ \sin x \approx x $

(b) Use a graphing device to determine the value of $ x $ for which $ \sin x $ and $ x $ differ by converting from radians to degrees.

Period of a pendulum A standard pendulum of length $L$ swing ing under only the influence of gravity (no resistance) has a period of

$$

T=\frac{4}{\omega} \int_{0}^{\pi / 2} \frac{d \varphi}{\sqrt{1-k^{2} \sin ^{2} \varphi}}

$$

where $\omega^{2}=g / L, k^{2}=\sin ^{2}\left(\theta_{0} / 2\right), g=9.8 \mathrm{m} / \mathrm{s}^{2}$ is the

acceleration due to gravity, and $\theta_{0}$ is the initial angle from whic the pendulum is released (in radians). Use numerical integration to approximate the period of a pendulum with $L=1 \mathrm{m}$ that is released from an angle of $\theta_{0}=\pi / 4$ rad.

The period of a simple pendulum, defined as the time necessary for one complete oscillation, is measured in time units and is given by the equation $$T=2 \pi \sqrt{\frac{L}{a_{g}}}$$ where $L$ is the length of the pendulum and $a_{g}$ is the acceleration due to gravity, which has units of length divided by time squared. Check this equation for dimensional consistency.

Period of a Pendulum The period $T$ (in seconds) of a simple pendulum is a function of its length $l$ (in feet) defined by the equation

$$T=2 \pi \sqrt{\frac{l}{g}}$$

where $g \approx 32.2$ feet per second per second is the acceleration of gravity.

(a) Use a graphing utility to graph the function $T=T(l)$

(b) Now graph the functions $T=T(l+1), T=T(l+2)$ and $T=T(l+3)$

(c) Discuss how adding to the length $l$ changes the period $T$.

(d) Now graph the functions $T=T(2 l), T=T(3 l),$ and $T=T(4 l)$

(e) Discuss how multiplying the length $l$ by factors of $2,3$ and 4 changes the period $T$.

The swinging pendulum Mrs. Hanrahan’s precalculus class collected data on the length (in

centimeters) of a pendulum and the time (in seconds) the pendulum took to complete one back-and-forth swing (called its period). Here are their data:

(a) Make a reasonably accurate scatterplot of the data by hand, using length as the explanatory variable.

Describe what you see.

(b) The theoretical relationship between a pendulum’s length and its period is

$$=\frac{2 \pi}{\sqrt{g}} \sqrt{\text { length }}$$

where $g$ is a constant representing the acceleration due to gravity (in this case, $g=980 {cm} / {s}^{2} )$ . Use the graph below to identify the transformation that was

used to linearize the curved pattern in part (a).

(c) Use the following graph to identify the transformation that was used to linearize the curved pattern in part (a).

The simple pendulum can be thought of as a special case of the physical pendulum where all of the mass is at a distance $L$ from the rotation axis. For a simple pendulum of mass $m$ and length $L$, show that the expression for the period of a physical pendulum (Eq. $10-27$ ) reduces to the expression for the period of a simple pendulum (Eq. $10-26 b$ ).

Pendulum with Varying Length. A pendulum is formed by a mass m attached to the end of a wire that is attached to the ceiling. Assume that the length $I(t)$ of the wire varies with time in some predetermined fashion. If $\theta(t)$ is the angle in radians between the pendulum and the vertical, then the motion of the pendulum is governed for small angles by the initial value problem

$$\begin{array}{l}{l^{2}(t) \theta^{\prime \prime}(t)+2 l(t) l^{\prime}(t) \theta^{\prime}(t)+g l(t) \sin (\theta(t))=0} \\ {\theta(0)=\theta_{0}, \quad \theta^{\prime}(0)=\theta_{1}}\end{array}$$

where g is the acceleration due to gravity. Assume that

$$l(t)=I_{0}+l_{1} \cos (\omega t-\phi)$$

where $l_{1}$ is much smaller than $l_{0}$ l0. (This might be a model

for a person on a swing, where the $pumping$ action

changes the distance from the center of mass of the swing to the point where the swing is attached.) To simplify the computations, take $g=1$Using the Runge-Kutta algorithm with $h=0.1$ study the motion of the pendulum when $\theta_{0}=0.05, \theta_{1}=0, \quad l_{0}=1, l_{1}=0.1$ $\omega=1,$ and $\phi=0.02 .$ In particular, does the pendulum

ever attain an angle greater in absolute value than the

initial angle $\theta_{0} ?$

A small object is attached to the end of a string to form a simple pendulum. The period of its harmonic motion is measured for small angular displacements and three lengths. For lengths of $1.000 \mathrm{m}, 0.750 \mathrm{m},$ and $0.500 \mathrm{m},$ total time intervals for 50 oscillations of $99.8 \mathrm{s}, 86.6 \mathrm{s}$, and $71.1 \mathrm{s}$ are measured with a stopwatch. (a) Determine the period of motion for each length. (b) Determine the mean value of $g$ obtained from these three independent measurements and compare it with the accepted value. (c) Plot $T^{2}$ versus $L$ and obtain a value for $g$ from the slope of your best-fit straightline graph. (d) Compare the value found in part (c) with that obtained in part (b).

$21-28=$ Damped Harmonic Motion An initial amplitude $k$ ,

damping constant $c$ , and frequency $f$ or period $p$ are given. (Recall

that frequency and period are related by the equation $f=1 / p$ .

$$

\begin{array}{l}{\text { (a) Find a function that models the damped harmonic }} \\ {\text { motion. Use a function of the form } y=k e^{-c t} \cos \omega t \text { in }} \\ {\text { Exercises } 21-24 \text { and of the form } y=k e^{-c t} \sin \omega t \text { in }} \\ {\text { Exercises } 25-28 \text { . }} \\ {\text { (b) Graph the function. }}\end{array}

$$

$$

k=15, \quad c=0.25, \quad f=0.6

$$

Using the vectorized Runge-Kutta algorithm for systems with $h=0.125$ approximate the solution to the initial value problem

$$\begin{array}{ll}{x^{\prime}=2 x-y ;} & {x(0)=0} \\ {y^{\prime}=3 x+6 y ;} & {y(0)=-2}\end{array}$$

at $t=1$ Compare this approximation to the actual solution

$$x(t)=e^{5 t}-e^{3 t}, \quad y(t)=e^{3 t}-3 e^{5 t}$$

Length of a Pendulum A simple pendulum swings back and forth in regular time intervals. Grandfather clocks use pendulums to keep accurate time. The relationship between the length

of a pendulum $L$ and the period (time) $T$ for one complete oscil-lation can be expressed by the function $L=k T^{n},$ where $k$ is a constant and $n$ is a positive integer to be determined. The data below were taken for different lengths of pendulums.

Source: Gary Rockswold.

(a) Find the value of $k$ for $n=1,2,$ and $3,$ using the data forthe 4 -ft pendulum.

(b) Use a graphing calculator to plot the data in the table and to graph the function $L=k T^{n}$ for the three values of $k$ (and their corresponding values of $n )$ found in part (a). Which function best fits the data?

(c) Use the best-fitting function from part (a) to predict the period of a pendulum having a length of 5 $\mathrm{ft.}$

(d) If the length of pendulum doubles, what happens to the period?

(e) If you have a graphing calculator or computer program with a quadratic regression feature, use it to find a quadratic function that approximately fits the data. How does this answer compare with the answer to part (b)?

A pendulum is made from a uniform rod of mass $m_{1}$ and a small block of mass $m_{2}$ attached at the lower end. (a) If the length of the pendulum is $L$ and the oscillations are small, find the period of the oscillations in terms of $m_{1}, m_{2}, L,$ and $g .$ (b) Check your answer to part (a) in the two special cases $m_{1} \gg m_{2}$ and $m_{1}<<m_{2}$.

The angular position of a pendulum is represented by the equation $\theta=0.0320 \cos \omega t,$ where $\theta$ is in radians and $\omega=$

$4.43 \mathrm{rad} / \mathrm{s} .$ Determine the period and length of the pendulum.

$21-28=$ Damped Harmonic Motion An initial amplitude $k$ ,

damping constant $c$ , and frequency $f$ or period $p$ are given. (Recall

that frequency and period are related by the equation $f=1 / p$ .

$$

\begin{array}{l}{\text { (a) Find a function that models the damped harmonic }} \\ {\text { motion. Use a function of the form } y=k e^{-c t} \cos \omega t \text { in }} \\ {\text { Exercises } 21-24 \text { and of the form } y=k e^{-c t} \sin \omega t \text { in }} \\ {\text { Exercises } 25-28 \text { . }} \\ {\text { (b) Graph the function. }}\end{array}

$$

$$

k=100, \quad c=0.05, \quad p=4

$$

$$

\begin{array}{l}{\text { For an object in simple harmonic motion with amplitude } a} \\ {\text { and period } 2 \pi / \omega, \text { find an equation that models the displace- }} \\ {\text { ment } y \text { at time } t \text { if }}\end{array}

$$

$$

\text { (a) } y=0 \text { at time } t=0: y=

$$

$$

\text { (b) } y=a \text { at time } t=0: y=

$$

A pendulum, consisting of a bob of mass $M$ on a cord of length $L$, is interrupted in its swing by a peg a distance $d$ below its point of suspension. (a) If the bob is to travel in a full circle of radius $(L-d)$ around the peg, what is the minimum possible speed it can have at the lowest point in its motion, just before it starts to go around? Ignore any decrease in the length of the string due to the peg's circumference. (b) From what minimum angle $\theta$ must the pendulum be released so that the bob attains the speed calculated in (a)?

CAN'T COPY THE FIGURE

Write an equation that relates the quantities.

$$

\begin{array}{l}{\text { The period of a pendulum is the }} \\ {\text { time required for one oscillation; the pendulum is usually }} \\ {\text { referred to as simple when the angle made to the vertical is }} \\ {\text { less than } 5^{\circ} . \text { The period } T \text { of a simple pendulum (in seconds) }} \\ {\text { varies directly with the square root of its length } l \text { (in feet). }} \\ {\text { The constant of proportionality is } \frac{2 \pi}{\sqrt{32}} \text { . }}\end{array}

$$

The time $T$ (in seconds) required for a pendulum to make one complete swing back and forth is approximated by

$$

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

$$

where $g$ is the acceleration due to gravity and $L$ is the length of the pendulum (in feet).

a. Find the length of a pendulum that requires 1.36 sec to make one complete swing back and forth. (Assume that the acceleration due to gravity is $g=32$ ft $/$ sec $^{2}$. Round to the nearest tenth of a foot.

b. Find the time required for a pendulum to complete one swing back and forth if the length of the pendulum is $4 \mathrm{ft}$. (Assume that the acceleration due to gravity is $g=32 \mathrm{ft} / \mathrm{sec}^{2}$ ) Round to the nearest tenth of a second.

$21-28=$ Damped Harmonic Motion An initial amplitude $k$ ,

damping constant $c$ , and frequency $f$ or period $p$ are given. (Recall

that frequency and period are related by the equation $f=1 / p$ .

$$

\begin{array}{l}{\text { (a) Find a function that models the damped harmonic }} \\ {\text { motion. Use a function of the form } y=k e^{-c t} \cos \omega t \text { in }} \\ {\text { Exercises } 21-24 \text { and of the form } y=k e^{-c t} \sin \omega t \text { in }} \\ {\text { Exercises } 25-28 \text { . }} \\ {\text { (b) Graph the function. }}\end{array}

$$

$$

k=1, \quad c=1, \quad p=1

$$

Clock Pendulum The pendulum in a grandfather clock makes

one complete swing every 2 s. The maximum angle that the pendulum makes with respect to its rest position is $10^{\circ} .$ We know

from physical principles that the angle $\theta$ between the pendulum and its rest position changes in simple harmonic fashion. Find an

equation that describes the size of the angle $\theta$ as a function of

time. (Take $t=0$ to be a time when the pendulum is vertical.)

damping constant $c$ , and frequency $f$ or period $p$ are given. (Recall

that frequency and period are related by the equation $f=1 / p$ .

$$

$$

k=7, \quad c=10, \quad p=\pi / 6

$$

Deal with the simple pendulum whose motion is described by Equation ( 8.5 .28 ).

Show that the period of the simple pendulum is $T=$ $2 \pi \sqrt{L / g} .$ Determine the length of a pendulum that takes one second to swing from its extreme position on the right to its extreme position on the left. Let $g=9.8$ meters/second $^{2}$

In quantum mechanics, the study of the Schrodinger equation for the case of a harmonic oscillator leads to a consideration of Hermite's equation, $\quad y^{\prime \prime}-2 t y^{\prime}+\lambda y=0$

where $\lambda$ is a parameter. Use the reduction of a second linearly independent solution to Hermite's equation for the given value of $\lambda$ and corresponding solution $f(t).$

$\begin{array}{ll}{\text { (a) } \lambda=4,} & {f(t)=1-2 t^{2}} \\ {\text { (b) } \lambda=6,} & {f(t)=3 t-2 t^{3}}\end{array}$

damping constant $c$ , and frequency $f$ or period $p$ are given. (Recall

that frequency and period are related by the equation $f=1 / p$ .

$$

$$

k=0.75, \quad c=3, \quad p=3 \pi

$$

damping constant $c$ , and frequency $f$ or period $p$ are given. (Recall

that frequency and period are related by the equation $f=1 / p$ .

$$

$$

k=2, \quad c=1.5, \quad f=3

$$

damping constant $c$ , and frequency $f$ or period $p$ are given. (Recall

that frequency and period are related by the equation $f=1 / p$ .

$$

$$

k=0.3, \quad c=0.2, \quad f=20

$$

Use the fourth-order Runge-Kutta subroutine with h = 0.25 to approximate the solution to the initial value problem

$$ y^{\prime}=2 y-6, \quad y(0)=1 $$

at x = 1. (Thus, input N = 4.) Compare this approximation to the actual solution $ y=3-2 e^{2 x} $ evaluated at x = 1.

Consider the damped forced motion described by

$$\frac{d^{2} y}{d t^{2}}+\frac{c}{m} \frac{d y}{d t}+\frac{k}{m} y=\frac{F_{0}}{m} \cos \omega t$$

We have shown that the steady-state solution can be written in the form

$$y_{p}(t)=\frac{F_{0}}{H} \cos (\omega t-v)$$

where

$$\begin{array}{c}

\cos v=\frac{m\left(\omega_{0}^{2}-\omega^{2}\right)}{H}, \quad \sin v=\frac{c \omega}{H} \\

\omega_{0}=\sqrt{\frac{k}{m}} \\

H=\sqrt{m^{2}\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+c^{2} \omega^{2}}

\end{array}$$

Assuming that $c^{2} /\left(2 m^{2} \omega_{0}^{2}\right)<1,$ show that the amplitude of the steady-state solution is a maximum when

$$\omega=\sqrt{\omega_{0}^{2}-\frac{c^{2}}{2 m^{2}}}$$

[Hint: The maximum occurs at the value of $\omega$ that makes $H$ a minimum. Assume that $H$ is a function of

$\omega, \text { and determine the value of } \omega \text { that minimizes } H .]$

damping constant $c$ , and frequency $f$ or period $p$ are given. (Recall

that frequency and period are related by the equation $f=1 / p$ .

$$

$$

k=12, \quad c=0.01, \quad f=8

$$

The two pendulums shown in $\textbf{Fig. E14.57}$ each consist of a uniform solid ball of mass $M$ supported by a rigid massless rod, but the ball for pendulum $A$ is very tiny while the ball for pendulum $B$ is much larger. Find the period of each pendulum for small displacements. Which ball takes longer to complete a swing?

(II) A simple pendulum is 0.30 $\mathrm{m}$ long. At $t=0$ it is released from rest starting at an angle of $13^{\circ} .$ Ignoring friction, what will be angular position of the pendulum at (a) $t=0.35 \mathrm{s},$ (b) $t=3.45 \mathrm{s},$ and $(c) t=6.00 \mathrm{s?}$

Temperature and the period of a pendulum For oscillations

of small amplitude (short swings), we may safely model the relationship between the period $T$ and the length $L$ of a simple pendulum with the equation

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

where $g$ is the constant acceleration of gravity at the pendulum's

location. If we measure $g$ in centimeters per second squared,

we measure $L$ in centimeters and $T$ in seconds. If the pendulum

is made of metal, its length will vary with temperature, either

increasing or decreasing at a rate that is roughly proportional to

L. In symbols, with $u$ being temperature and $k$ the proportionality

constant,

$$\frac{d L}{d u}=k L$$

Assuming this to be the case, show that the rate at which the period changes with respect to temperature is $k T / 2 .$

Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period $T$ and the length $L$ of a simple pendulum with the equation

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

where $g$ is the constant acceleration of gravity at the pendulum's location. If we measure $g$ in centimeters per second squared, we measure $L$ in centimeters and $T$ in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to $L$. In symbols, with $u$ being temperature and $k$ the proportionality constant,

$$\frac{d L}{d u}=k L$$

Assuming this to be the case, show that the rate at which the period changes with respect to temperature is $k T / 2$

A simple pendulum consists of a ball of mass 5.00 $\mathrm{kg}$ hanging from a uniform string of mass 0.0600 $\mathrm{kg}$ and length $L$ . If the period of oscillation of the pendulum is 2.00 $\mathrm{s}$ s, determine the speed of a transverse wave in the string when the pendulum hangs vertically.