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Chapter 11: Problem 33

A "seconds pendulum" beats seconds; that is, it takes 1 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 Pendulum Formula

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

Solve for Length of the 'Seconds Pendulum'

For a seconds pendulum, it takes 1 second for half a cycle, meaning 2 seconds for a full cycle (\( T = 2 \) s). Using the formula, re-arrange to find \( L \):\[ L = \frac{gT^2}{4\pi^2} \]Substitute \( g = 9.80 \, \mathrm{m/s^2} \) and \( T = 2 \, \mathrm{s} \):\[ L = \frac{9.80 \times 2^2}{4\times\pi^2} \approx 0.994 \, \mathrm{m} \].
03

Solve for Length with T = 1.00 s

Now, find the length of the pendulum with a period \( T = 1 \, \mathrm{s} \). Using the same rearranged formula \[ L = \frac{gT^2}{4\pi^2} \]Substitute \( g = 9.80 \, \mathrm{m/s^2} \) and \( T = 1 \, \mathrm{s} \):\[ L = \frac{9.80 \times 1^2}{4\times\pi^2} \approx 0.248 \, \mathrm{m} \].

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Key Concepts

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

Seconds Pendulum
A "seconds pendulum" is a special type of pendulum that has a period of precisely two seconds. This means it takes one second to swing in one direction and another second to return, making its total cycle two seconds. Due to this characteristic, the seconds pendulum is a valuable tool for timing and is often mentioned in discussions on classic mechanics regarding pendulum motion.

Understanding how the seconds pendulum works is crucial in various scientific and engineering applications. The length of a seconds pendulum is directly related to the acceleration due to gravity in its location. This relationship helps in understanding both time measurement and the local effects of gravity on pendulum mechanics.
Pendulum Period Calculation
Calculating the period of a pendulum involves determining how long it takes to complete one full swing, going both to and fro. The formula for the period of a pendulum is given by \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( T \) represents the period in seconds, \( L \) is the pendulum length in meters, and \( g \) is the acceleration due to gravity in meters per second squared.

To calculate the period, you need to know the length of the pendulum as well as the acceleration due to gravity in the area where the pendulum is located. By rearranging this formula, you can also solve for the pendulum's length if you have the other two variables. This is helpful in situations where you wish to design a pendulum with a specific swinging period.
Acceleration Due to Gravity
The acceleration due to gravity, denoted as \( g \), is a key factor in pendulum calculations and varies depending on location. On Earth, \( g \) is typically taken as \( 9.80 \mathrm{\, m/s^2} \) unless otherwise specified. This value represents the rate at which objects accelerate towards the Earth's surface.

In pendulum experiments, \( g \) affects how fast a pendulum swings. Different values of \( g \) would result in different swinging behaviors for the same pendulum length. Understanding and utilizing this gravitational acceleration is important to predict pendulum motion accurately in different environments, such as on other planets or in varying geographical locations on Earth.
Simple Pendulum Formula
The simple pendulum formula is immensely helpful for predicting and analyzing the motion of pendulums. This formula, \( T = 2\pi \sqrt{\frac{L}{g}} \), is derived under the assumption that the pendulum behaves in simple harmonic motion, where the only force acting on it is gravity. Simple harmonic motion means a repetitive movement back and forth through an equilibrium, or central, position.

When using the simple pendulum formula, it's essential to ensure that certain conditions are met鈥攑rimarily, the angle of swing needs to be relatively small. In such conditions, the formula can predict the period with high accuracy. This formula forms the basis of many real-world applications, from traditional clocks to modern-day engineering designs. Understanding it allows us to apply basic physics principles to solve complex problems in kinematics and dynamics.

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Most popular questions from this chapter

A \(50-\mathrm{g}\) mass vibrates in SHM at the end of a spring. The amplitude of the motion is \(12 \mathrm{~cm}\), and the period is \(1.70 \mathrm{~s}\). Find: \((a)\) the frequency, \((b)\) the spring constant, \((c)\) the maximum speed of the mass, \((d)\) the maximum acceleration of the mass, \((e)\) the speed when the displacement is \(6.0 \mathrm{~cm}\), and \((f)\) the acceleration when \(x=6.0 \mathrm{~cm}\). (a) \(f=\frac{1}{T}=\frac{1}{1.70 \mathrm{~s}}=0.588 \mathrm{~Hz}\) (b) Since \(T=2 \pi \sqrt{m / k}\) $$ k=\frac{4 \pi^{2} m}{T^{2}}=\frac{4 \pi^{2}(0.050 \mathrm{~kg})}{(1.70 \mathrm{~s})^{2}}=0.68 \mathrm{~N} / \mathrm{m} $$ (c) \(\left|v_{0}\right|=x_{0} \sqrt{\frac{k}{m}}=(0.12 \mathrm{~m}) \sqrt{\frac{0.68 \mathrm{~N} / \mathrm{m}}{0.050 \mathrm{~kg}}}=0.44 \mathrm{~m} / \mathrm{s}\) (d) From \(a=-(k / m) x\) it is seen that \(a\) has maximum magnitude when \(x\) has maximum magnitude, that is, at the endpoints \(x=\pm x_{0}\). Thus, $$ a_{0}=\frac{k}{m} x_{0}=\frac{0.68 \mathrm{~N} / \mathrm{m}}{0.050 \mathrm{~kg}}(0.12 \mathrm{~m})=1.6 \mathrm{~m} / \mathrm{s}^{2} $$ (e) From \(|v|=\sqrt{\left(x_{0}^{2}-x^{2}\right)(k / m)}\), $$ |v|=\sqrt{\frac{\left[(0.12 \mathrm{~m})^{2}-(0.06 \mathrm{~m})^{2}\right](0.68 \mathrm{~N} / \mathrm{m})}{(0.050 \mathrm{~kg})}}=0.38 \mathrm{~m} / \mathrm{s} $$ (f) \(a=-\frac{k}{m} x=-\frac{0.68 \mathrm{~N} / \mathrm{m}}{0.050 \mathrm{~kg}}(0.060 \mathrm{~m})=-0.82 \mathrm{~m} / \mathrm{s}^{2}\)

Find the frequency of vibration on Mars for a simple pendulum that is \(50 \mathrm{~cm}\) long. Objects weigh \(0.40\) as much on Mars as on the Earth.

With a \(50-g\) mass at its end, a spring undergoes SHM with a frequency of \(0.70 \mathrm{~Hz}\). How much work is done in stretching the spring \(15 \mathrm{~cm}\) from its unstretched length? How much energy is then stored in the spring?

A particle that is at the origin of coordinates at exactly \(t=0\) vibrates about the origin along the \(y\) -axis with a frequency of \(20 \mathrm{~Hz}\) and an amplitude of \(3.0 \mathrm{~cm}\). Write out its equation of motion in centimeters.

A \(200-\mathrm{g}\) mass vibrates horizontally without friction at the end of a horizontal spring for which \(k=7.0 \mathrm{~N} / \mathrm{m} .\) The mass is displaced \(5.0 \mathrm{~cm}\) from equilibrium and released. Find \((a)\) its maximum speed and \((b)\) its speed when it is \(3.0 \mathrm{~cm}\) from equilibrium. \((c)\) What is its acceleration in each of these cases? From the conservation of energy, $$ \frac{1}{2} k x_{0}^{2}=\frac{1}{2} m v^{2}+\frac{1}{2} k x^{2} $$ where \(k=7.0 \mathrm{~N} / \mathrm{m}, x_{0}=0.050 \mathrm{~m}\), and \(m=0.200 \mathrm{~kg}\). Solving for \(|v|\) gives $$ |v|=\sqrt{\left(x_{0}^{2}-x^{2}\right) \frac{k}{m}} $$ (a) The speed is a maximum when \(x=0 ;\) that is, when the mass is passing through the equilibrium position: $$ |v|=x_{0} \sqrt{\frac{k}{m}}=(0.050 \mathrm{~m}) \sqrt{\frac{7.0 \mathrm{~N} / \mathrm{m}}{0.200 \mathrm{~kg}}}=0.30 \mathrm{~m} / \mathrm{s} $$ (b) When \(x=0.030 \mathrm{~m}\), $$ |v|=\sqrt{\frac{7.0 \mathrm{~N} / \mathrm{m}}{0.200 \mathrm{~kg}}\left[(0.050)^{2}-(0.030)^{2}\right] \mathrm{m}^{2}}=0.24 \mathrm{~m} / \mathrm{s} $$ (c) Using \(F=m a\) and \(F=k x\), $$ a=\frac{k}{m} x=\left(35 \mathrm{~s}^{-2}\right)(x) $$ which yields \(a=0\) when the mass is at \(x=0\), and \(a=1.1 \mathrm{~m} / \mathrm{s}^{2}\) when \(x=0.030 \mathrm{~m}\).
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