Oscillatory Motion and Waves and Physics of Hearing.
119 The Simple Pendulum
Learning Objectives
- Measure the acceleration due to gravity using a simple pendulum.
A simple pendulum consists of a small, dense mass—called the bob—attached to a lightweight string or wire, as shown in Figure 119.1. For small displacements, such a pendulum undergoes simple harmonic motion. Pendulums are widely used in everyday life—ranging from timekeeping in clocks to swings in playgrounds.
The displacement of the pendulum is defined as the arc length [latex]s[/latex] from the equilibrium position. The net force acting on the bob is tangent to this arc and given by:
where [latex]m[/latex] is the mass of the bob, [latex]g[/latex] is the gravitational acceleration, and [latex]\theta[/latex] is the angular displacement from vertical. The component of the gravitational force along the arc causes the bob to accelerate back toward equilibrium.
To determine if the pendulum exhibits simple harmonic motion, we examine the force behavior for small angles. For displacements less than approximately [latex]15^\circ[/latex], we can use the approximation:
when [latex]\theta[/latex] is measured in radians. Thus, the restoring force becomes:
We also relate the arc length to the angle via:
Substituting back into the force equation gives:
This has the form of Hooke’s Law:
with effective spring constant [latex]k = \frac{mg}{L}[/latex] and displacement [latex]x = s[/latex]. Therefore, for small displacements, the simple pendulum behaves as a simple harmonic oscillator.
We can now derive the period [latex]T[/latex] of a pendulum. For a simple harmonic oscillator:
Substituting [latex]k = \frac{mg}{L}[/latex] gives:
This final expression shows that the period of a simple pendulum depends only on the pendulum’s length [latex]L[/latex] and the local acceleration due to gravity [latex]g[/latex]. It is independent of mass and—provided that [latex]\theta[/latex] remains small—mostly independent of amplitude.
This result has important applications. For instance, by measuring the period and length of a pendulum, one can determine the value of [latex]g[/latex]. In fact, precise pendulum measurements were historically used to determine gravitational acceleration at various locations on Earth.
Example 119.1: Measuring Acceleration due to Gravity: The Period of a Pendulum
What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?
Strategy
We are asked to find [latex]g[/latex] given the period [latex]T[/latex] and the length [latex]L[/latex] of a pendulum. We can solve [latex]T=2\pi \sqrt{\frac{L}{g}}[/latex] for [latex]g[/latex], assuming only that the angle of deflection is less than [latex]\text{15º}[/latex].
Solution
- Square [latex]T=2\pi \sqrt{\frac{L}{g}}[/latex] and solve for [latex]g[/latex]:
[latex]g={4\pi }^{2}\frac{L}{{T}^{2}}.[/latex]
- Substitute known values into the new equation:
[latex]g={4\pi }^{2}\frac{0\text{.}\text{75000}\phantom{\rule{0.25em}{0ex}}\text{m}}{{\left(1\text{.}\text{7357 s}\right)}^{2}}.[/latex]
- Calculate to find [latex]g[/latex]:
[latex]g=9\text{.}\text{8281}\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{2}.[/latex]
Discussion
This method for determining [latex]g[/latex] can be very accurate. This is why length and period are given to five digits in this example. For the precision of the approximation [latex]\text{sin θ}\approx \theta[/latex] to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about [latex]\text{0.5º}[/latex].
Making Career Connections
Knowing the local value of [latex]g[/latex] (acceleration due to gravity) can be critical in fields such as geological exploration. For example, creating a detailed map of variations in [latex]g[/latex] over a geographic region can assist in the study of tectonic plate activity or in identifying subsurface resources like oil fields and mineral deposits.
Take-Home Experiment: Determining [latex]g[/latex]
Try measuring [latex]g[/latex] in your own location using a simple pendulum. Cut a string (or dental floss) to a length of about 1 meter. Tie a dense object (like a metal nut or car key) to one end. Displace the pendulum by less than [latex]10^\circ[/latex] and let it swing freely. Use a stopwatch to time 10 full oscillations and calculate the period. Then determine [latex]g[/latex] using the formula:
How accurate is your result? What are some ways to reduce error or improve the precision of your measurement?
Check Your Understanding
An engineer constructs two simple pendula. Each is suspended by a wire from the same ceiling, and both hang 2 cm above the floor. Pendulum 1 has a bob with mass [latex]10\ \text{kg}[/latex], and Pendulum 2 has a bob with mass [latex]100\ \text{kg}[/latex]. If both are displaced by [latex]12^\circ[/latex], how will their motion differ?
They will behave identically. The period of a simple pendulum is independent of its mass and depends only on the length and gravitational acceleration. Thus, both pendula will oscillate with the same motion.
PhET Explorations: Pendulum Lab
Use this interactive simulation to experiment with one or two pendula. Explore how changing the string length, mass, and amplitude affects the pendulum’s period. Use tools like the photogate timer to accurately measure timing. You can also modify friction or simulate conditions on different planets. Try finding [latex]g[/latex] on planet X and observe how motion becomes non-harmonic at large amplitudes.
Section Summary
- A mass [latex]m[/latex] suspended by a string of length [latex]L[/latex] forms a simple pendulum. For displacements less than about [latex]15^\circ[/latex], the motion is simple harmonic.
- The period [latex]T[/latex] of a simple pendulum is given by:
[latex]T = 2\pi \sqrt{\frac{L}{g}}[/latex]
where [latex]L[/latex] is the pendulum length and [latex]g[/latex] is the acceleration due to gravity.
Conceptual Questions
- Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.
Problems & Exercises
As usual, the acceleration due to gravity in these problems is taken to be[latex]g=9.80\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{2}[/latex], unless otherwise specified.
- What is the length of a pendulum that has a period of 0.500 s?
- Some people think a pendulum with a period of 1.00 s can be driven with “mental energy” or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum?
- What is the period of a 1.00-m-long pendulum?
- How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?
- The pendulum on a cuckoo clock is 5.00 cm long. What is its frequency?
- Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?
- (a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is [latex]9\text{.}\text{79}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}[/latex] is moved to a location where it the acceleration due to gravity is [latex]9\text{.}\text{82}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}[/latex]. What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity.
- A pendulum with a period of 2.00000 s in one location [latex]\left(g=9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)[/latex] is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?
- (a) What is the effect on the period of a pendulum if you double its length? (b) What is the effect on the period of a pendulum if you decrease its length by 5.00%?
- Find the ratio of the new/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is [latex]1\text{.}\text{63}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}[/latex].
- At what rate will a pendulum clock run on the Moon, where the acceleration due to gravity is [latex]1\text{.}\text{63}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}[/latex], if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock’s hour hand to make one revolution on the Moon.
- Suppose the length of a clock’s pendulum is changed by 1.000%, exactly at noon one day. What time will it read 24.00 hours later, assuming it the pendulum has kept perfect time before the change? Note that there are two answers, and perform the calculation to four-digit precision.
- If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect time?
Glossary
- simple pendulum
- an object with a small mass suspended from a light wire or string
an object with a small mass suspended from a light wire or string
the oscillatory motion in a system where the net force can be described by Hooke’s law
