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(II) A pendulum consists of a tiny bob of mass \(M\) and a uniform cord of mass \(m\) and length \(\ell\) (a) Determine a formula for the period using the small angle approximation. (b) What would be the fractional error if you use the formula for a simple pendulum, Eq. 12 \(\mathrm{c} ?\) $$T=\frac{1}{f}=2 \pi \sqrt{\frac{\ell}{g}} \quad[\theta\( small \)] \quad(12 \mathrm{c})$$

Short Answer

Expert verified
(a) T = 2\(\pi \) \(\\sqrt{\left (\frac{\ell \cdot \frac{M + \frac{m}{2}}{M + m}}{g} \right)}\). (b) Fractional Error = \(\frac{1}{2} \frac{m}{M}\).

Step by step solution

01

Understanding the Small Angle Approximation

In this problem, we need to determine the period of a pendulum, which consists of two parts: a bob and a cord of uniform mass using the small angle approximation. Using this approximation simplifies the calculations as it assumes the angle is close to zero, allowing us to treat sin(θ) as approximately equal to θ.
02

Construct the Physical Model

For a pendulum with a light cord and a heavy bob, the period depends primarily on the length and effective force of gravity acting on it. Because the cord has mass, we need to consider the distribution of this mass along with the bob's mass, which affects the moment of inertia and hence the pendulum's dynamics.
03

Identify Key Components' Contribution

The system consists of a bob of mass M and a cord of mass m and length \(\ell\). The effective length and mass need to include both the bob's mass and the cord's distributed mass. The center of mass for the cord is located at \(\frac{\ell}{2}\), thus influencing the period.
04

Derive Period Formula Using Reduced Mass

Incorporating the distributed mass of the string, the effective pendulum length is given by:\[L = \ell \cdot \frac{M + \frac{m}{2}}{M + m}\] The effective gravitational force is calculated using this effective length in the standard period formula for small angles: \[T = 2\pi \sqrt{\frac{L}{g}}\]
05

Calculate Fractional Error

The fractional error is calculated by comparing the derived period formula for the actual pendulum against the simple pendulum formula given in Eq. 12\(c\):\[\text{Fractional Error} = \frac{T_{actual} - T_{simple}}{T_{simple}} = \frac{2\pi \sqrt{\frac{\ell \cdot \frac{M + \frac{m}{2}}{M + m}}{g}} - 2\pi \sqrt{\frac{\ell}{g}}}{2\pi \sqrt{\frac{\ell}{g}}}\] This results in:\[\text{Fractional Error} = \frac{1}{2} \frac{m}{M} \] when expanded to first order for small \(\frac{m}{M}\).

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Key Concepts

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

Simple Pendulum
A simple pendulum is a basic mechanical system that consists of a small weight, usually referred to as the 'bob,' attached to a string or a rod of fixed length. Its motion involves swinging back and forth under the influence of gravity. This swinging motion is what we call oscillation. The most interesting characteristic of a simple pendulum is its period, which is the time it takes to complete one full swing back and forth.

The formula for the period of a simple pendulum is given by \[T = 2\pi \sqrt{\frac{\ell}{g}}\], where \(\ell\) is the length of the string and \(g\) is the acceleration due to gravity.**Key Features of a Simple Pendulum**
  • The period depends only on the length of the string and the acceleration due to gravity.
  • This formula is specifically valid when assuming the angle is small (small angle approximation).
When a pendulum is swinging with small amplitude, the motion is similar to simple harmonic motion, allowing us to use the simple pendulum formula accurately for such cases. Understanding the simple pendulum is fundamental for delving deeper into other more complex oscillatory systems.
Small Angle Approximation
The small angle approximation is a mathematical simplification often used in physics to make complex equations more manageable. It relies on the assumption that the angle \(\theta\) is small, or close to zero, measured in radians.**How It Works**
When \(\theta\) is small, the values of \(\sin(\theta)\), \(\tan(\theta)\), and \(\theta\) (in radians) are approximately equal. This simplification is valuable when solving problems involving pendulums because it allows us to assume that \[ \sin(\theta) \approx \theta.\] This assumption is integral as it simplifies the differential equations governing the motion of the pendulum.**Applications**
  • Useful in deriving the period formula of a simple pendulum and other oscillatory systems.
  • Significantly reduces the complexity of calculations.
As useful as this approximation is, it is essential to be cautious with its application. It is only valid for small angles, usually less than 15 degrees, where the error introduced is typically negligible. In cases of larger angles, the approximation can lead to significant errors.
Moment of Inertia
The moment of inertia is a physical property that measures an object's resistance to angular acceleration about a particular axis. It is analogous to mass in linear motion, playing a crucial role in rotational dynamics. For a pendulum system, understanding moment of inertia helps us appreciate how the distribution of mass affects the pendulum's period and behavior.**For a Pendulum**
When considering a pendulum, especially with a cord of non-negligible mass, the moment of inertia significantly influences the dynamics. In this context, you need to account for the distributed mass of the cord to accurately calculate the period of oscillation.**Key Considerations**
  • Higher moment of inertia means the system will require more torque to achieve the same angular acceleration.
  • In the exercise, the center of mass of the cord is positioned at \( \frac{\ell}{2} \), which impacts the effective length and, thus, the period.
By incorporating the moment of inertia into our model, we account for both the bob and the distributed mass of the cord. This results in a more accurate calculation of the pendulum's period, especially in systems where the string's mass cannot be ignored. Understanding these dynamics is crucial for solving complex problems involving rotational motion and oscillation.

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Most popular questions from this chapter

A small fly of mass \(0.25 \mathrm{~g}\) is caught in a spider's web. The web oscillates predominately with a frequency of \(4.0 \mathrm{~Hz}\). (a) What is the value of the effective spring stiffness constant \(k\) for the web? ( \(b\) ) At what frequency would you expect the web to oscillate if an insect of mass \(0.50 \mathrm{~g}\) were trapped?

(II) A 1.60 -kg object oscillates from a vertically hanging light spring once every 0.55 s. (a) Write down the equation giving its position \(y(+\) upward) as a function of time \(t,\) assuming it started by being compressed 16 \(\mathrm{cm}\) from the equilibrium position (where \(y=0 )\) , and released. \((b)\) How long will it take to get to the equilibrium position for the first time? (c) What will be its maximum speed? (d) What will be its maximum acceleration, and where will it first be attained?

An elastic cord is \(65 \mathrm{~cm}\) long when a weight of \(75 \mathrm{~N}\) hangs from it but is \(85 \mathrm{~cm}\) long when a weight of \(180 \mathrm{~N}\) hangs from it. What is the "spring" constant \(k\) of this elastic cord?

An oxygen atom at a particular site within a DNA molecule can be made to execute simple harmonic motion when illuminated by infrared light. The oxygen atom is bound with a spring-like chemical bond to a phosphorus atom, which is rigidly attached to the DNA backbone. The oscillation of the oxygen atom occurs with frequency \(f=3.7 \times 10^{13}\) . If the oxygen atom at this site is chemically replaced with a sulfur atom, the spring constant of the bond is unchanged (sulfur is just below oxygen in the Periodic Table). Predict the frequency for a DNA molecule after the sulfur substitution.

(II) What is the period of a simple pendulum 53 \(\mathrm{cm}\) long (a) on the Earth, and \((b)\) when it is in a freely falling elevator?

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