Question

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 due to gravity. (a) Use a graphing utility to graph the function \(T=T(l)\). (b) Now graph the functions \(T=T(l+1), T=T(l+2)\) $$ \text { and } T=T(l+3) $$ (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 \(T\)

Short answer

Adding to \( l \) increases \( T \). Multiplying \( l \) increases \( T \) but at a slower rate.

Step-by-step solution

Understand the given formula

The period of a simple pendulum is given by the formula: \[ T = 2 \pi \sqrt{\frac{l}{g}} \]where \( l \) is the length in feet and \( g \approx 32.2 \ \text{ft/s}^2 \) is the acceleration due to gravity.

Graph the function \( T=T(l) \)

To graph the function \( T=T(l) \), use a graphing utility. Set up the function \[ T = 2 \pi \sqrt{\frac{l}{32.2}} \] and plot it for a range of \( l \) values (e.g., from 0 to 10 feet).

Graph \( T=T(l+1) \), \( T=T(l+2) \), and \( T=T(l+3) \)

Modify the original function for different length values:For \( T=T(l+1) \), graph \[ T = 2 \pi \sqrt{\frac{l+1}{32.2}} \]For \( T=T(l+2) \), graph \[ T = 2 \pi \sqrt{\frac{l+2}{32.2}} \]For \( T=T(l+3) \), graph \[ T = 2 \pi \sqrt{\frac{l+3}{32.2}} \] using the same range of \( l \) values as before.

Discuss the effect of adding to the length \( l \)

Observe how the graphs change as \( l \) increases. Adding to the length \( l \) leads to an increase in the period \( T \). This is because the function \( T \) is directly proportional to \( \sqrt{l} \).

Graph \( T=T(2l) \), \( T=T(3l) \), and \( T=T(4l) \)

Modify the original function by multiplying \( l \) by factors of 2, 3, and 4:For \( T=T(2l) \), graph \[ T = 2 \pi \sqrt{\frac{2l}{32.2}} \]For \( T=T(3l) \), graph \[ T = 2 \pi \sqrt{\frac{3l}{32.2}} \]For \( T=T(4l) \), graph \[ T = 2 \pi \sqrt{\frac{4l}{32.2}} \] using the same range of \( l \) values.

Discuss the effect of multiplying the length \( l \)

Observe how the graphs change as \( l \) is multiplied by different factors. Multiplying the length \( l \) by 2, 3, or 4 significantly increases the period \( T \) since \( T \) is proportional to the square root of \( l \). This means the period increases but at a slower rate as \( l \) increases.

Key concepts

graphing utility

A graphing utility is a powerful tool that helps in visualizing mathematical functions. When graphing the pendulum period function, set up the equation:

\t
• \t\t\( T = 2 \pi \sqrt{\frac{l}{32.2}}\).\t

Choose a range for the length \( l \), for example from 0 to 10 feet, to see how the period \( T \) changes with different lengths. Use the graphing utility to plot this function. The curve will give us an insight into how the length of the pendulum affects its period.
When you graph functions like \( T = 2 \pi \sqrt{\frac{l+1}{32.2}} \), \( T = 2 \pi \sqrt{\frac{l+2}{32.2}} \), and \( T = 2 \pi \sqrt{\frac{l+3}{32.2}} \), you will observe how adding to the length impacts the period. Similarly, graphing \( T = 2 \pi \sqrt{\frac{2l}{32.2}} \), \( T = 2 \pi \sqrt{\frac{3l}{32.2}} \), and \( T = 2 \pi \sqrt{\frac{4l}{32.2}} \) demonstrates the effect of multiplying the length.

square root function

In the formula for the period of a simple pendulum, \( T = 2 \pi \sqrt{\frac{l}{g}} \), the square root function plays a crucial role. The square root function, \( \sqrt{x} \), is a mathematical operation that finds the number which, when multiplied by itself, yields the original number (in this case, the division of length \( l \) by the acceleration due to gravity \( g \)).
The square root function has unique properties:

\t
• It is always positive for positive input values.
\t
• It grows slower as the input value increases.

This means that even as the length of the pendulum increases, the period grows, but at a decreasing rate. This property is key to understanding why increasing the length of the pendulum results in a longer period of oscillation but not in a linear fashion.

acceleration due to gravity

The acceleration due to gravity, denoted as \( g \), is a constant that affects how objects fall. On Earth, \( g \approx 32.2 \text{ft/s}^2 \), meaning that in each second, an object's velocity increases by 32.2 feet per second due to gravity. In the equation \( T = 2 \pi \sqrt{\frac{l}{g}} \), \( g \) serves as the base measurement for the pendulum's responsiveness to gravity.
Because \( g \) is relatively constant on Earth's surface, it allows us to study the effects of other variables, such as the length of the pendulum, on its period. Variations in \( g \) would occur on different planetary bodies or if you measure at extreme altitudes. However, for standard physics problems involving pendulums, we assume \( g \) is constant.

simple pendulum equations

Simple pendulum equations help us understand the regular motion of a weight suspended from a pivot. The key equation for a simple pendulum's period is:

\t
• \t\t\( T = 2 \pi \sqrt{\frac{l}{g}} \)\t

This formula expresses the period \( T \) as a function of the pendulum’s length \( l \) and the acceleration due to gravity \( g \). It highlights how the period depends on the length of the pendulum.
By modifying this equation to study different scenarios:

\t
• \t\tAdding constants to the length (e.g., \( T = 2 \pi \sqrt{\frac{l+1}{g}} \)) shows how incremental increases lengthen the period.\t
\t
• \t\tMultiplying the length by constants (e.g., \( T = 2 \pi \sqrt{\frac{2l}{g}} \)) reveals how greater lengths affect the period more significantly.\t

Understanding these equations helps in predicting the behavior of pendulums in real-world applications. Studying them through graphing provides a visual representation, which simplifies complex concepts for better comprehension.