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A "seconds pendulum" beats seconds; that is, it takes \(1 \mathrm{~s}\) for half a cycle. (a) What is the length of a simple "seconds pendulum" at a place where \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} ?(b)\) What is the length there of \(a\) pendulum for which \(T=1.00 \mathrm{~s}\) ?

Short Answer

Expert verified
(a) 0.994 m (b) 0.248 m

Step by step solution

01

Understand the Simple Pendulum Formula

The formula for the period \( T \)of a simple pendulum is given by \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.
02

Identify Given Values

For part (a), we know that the period \( T = 2 \, \text{s} \) for a full cycle (as it beats seconds and takes 1 second for half a cycle). The gravitational acceleration \( g \) is given as \( 9.80 \, \text{m/s}^2 \).
03

Rearrange the Formula for Length

We need to solve for \( L \), so we rearrange the formula to \( L = \frac{T^2 g}{4\pi^2} \).
04

Calculate the Length for the Seconds Pendulum

Substitute \( T = 2 \, \text{s} \) and \( g = 9.80 \, \text{m/s}^2 \) into the rearranged formula: \[ L = \frac{(2)^2 \times 9.80}{4\pi^2} \].\Calculate to find \( L \).
05

Solve for Part (b) with T = 1.00 s

The problem states that for part (b), \( T = 1.00 \, \text{s} \). Use the same rearranged formula \( L = \frac{T^2 g}{4\pi^2} \).
06

Calculate the Length for T = 1.00 s

Substitute \( T = 1.00 \, \text{s} \) into the formula: \[ L = \frac{(1.00)^2 \times 9.80}{4\pi^2} \].\Calculate to find \( L \).
07

Interpret Results for Both Parts

After performing the calculations: \ (a) For the seconds pendulum, \( L \approx 0.994 \, \text{m} \). \ (b) For \( T = 1.00 \, \text{s} \), \( L \approx 0.248 \, \text{m} \). \These lengths agree with the expected behavior of pendulums of different periods.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pendulum Period
The pendulum period is the time it takes for one complete cycle of swinging back and forth. In the context of a simple pendulum, this period, denoted as \( T \), can be calculated using the formula:\[ T = 2\pi \sqrt{\frac{L}{g}} \]Here, \( T \) is the period, \( L \) is the pendulum length, and \( g \) is the gravitational acceleration. This equation tells us how long it takes for a pendulum to return to its starting position after a complete swing.
A pendulum that takes one second for a half cycle is called a "seconds pendulum," meaning it takes two seconds for a full cycle.
Understanding this concept helps us predict how changes in length and gravitational acceleration affect the pendulum's timing.
Gravitational Acceleration
Gravitational acceleration, represented by \( g \), is the rate at which an object accelerates when falling freely towards the Earth. Its standard value is approximately \( 9.81 \, \text{m/s}^2 \) but is often rounded to \( 9.80 \, \text{m/s}^2 \) for simplicity in calculations.In the case of a pendulum, \( g \) plays a crucial role in determining the pendulum's period. A higher gravitational acceleration indicates a stronger gravitational pull, which can make the pendulum swing faster, reducing its period. Conversely, a lower \( g \) slows down the pendulum, increasing the period.
Small differences in \( g \) at different locations on Earth can also result in slightly different pendulum behaviors.
Pendulum Length
Pendulum length, \( L \), is the distance from the pivot point to the center of mass (usually the "bob" at the end). This length directly influences the pendulum's period. Using the formula:\[ L = \frac{T^2 g}{4\pi^2} \]We can determine how the period and gravitational acceleration affect the length required to achieve a specific period.
For example, for a seconds pendulum at a location where \( g = 9.80 \, \text{m/s}^2 \), substituting the known values into the formula gives \( L \approx 0.994 \, \text{m} \).
If we're building a clock or timing device with pendulums, understanding how to calculate \( L \) allows us to ensure accurate timing by adjusting the pendulum's length appropriately.
Cycle Calculation
Cycle calculation involves determining the total time for a pendulum to complete a full swing, or cycle. This calculation uses the period \( T \) determined by various factors including the pendulum's length and gravitational acceleration.For a given \( T \), we can easily infer the length \( L \) using the rearranged period formula, provided \( g \) is known. For example, a pendulum with a period \( T = 1 \, \text{s} \) and \( g \) set to \( 9.80 \, \text{m/s}^2 \) uses the formula:\[ L = \frac{(1.00)^2 \times 9.80}{4\pi^2} \]Calculating this results in \( L \approx 0.248 \, \text{m} \).
Understanding and performing these cycle calculations are essential for applications where precise timekeeping based on pendulum swings is necessary.

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Most popular questions from this chapter

A 50 -g mass hangs at the end of a Hookean spring. When \(20 \mathrm{~g}\) more are added to the end of the spring, it stretches \(7.0 \mathrm{~cm}\) more. (a) Find the spring constant. ( \(b\) ) If the 20 -g mass is now removed, what will be the period of the motion? A 50 -g mass hangs at the end of a Hookean spring. When \(20 \mathrm{~g}\) more are added to the end of the spring, it stretches \(7.0 \mathrm{~cm}\) more. (a) Find the spring constant. ( \(b\) ) If the 20 -g mass is now removed, what will be the period of the motion?

How much energy is stored in a spring with an elastic constant of \(1000 \mathrm{~N} / \mathrm{m}\) when it is compressed \(10 \mathrm{~cm}\) ?

A particle vibrates according to the equation \(x=20 \cos 16 t\), where \(x\) is in centimeters. Find its amplitude, frequency, and position at exactly \(t=0 \mathrm{~s}\).

A particle oscillates according to the equation \(y=5.0 \cos 23 t\), where \(y\) is in centimeters. Find its frequency of oscillation and its position at \(t=0.15 \mathrm{~s}\).

A horizontal spring is set up like the one in \(\underline{\text { Fig. } 11-3 .}\) It has an elastic constant of \(50.0 \mathrm{~N} / \mathrm{m}\). A \(1.00\) -kg mass, sitting on a frictionless horizontal surface, is attached to the end of the spring. The mass is displaced \(10.0 \mathrm{~cm}\) to the right and released, whereupon it oscillates in SHM. Determine its acceleration (magnitude and direction) immediately after release. [Hint: \(x=x_{0}\) \(=10.0 \mathrm{~cm} .]\)

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