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Multiple-Concept Example 11 explores the concepts that are important in this problem. Pendulum A is a physical pendulum made from a thin, rigid, and uniform rod whose length is \(d .\) One end of this rod is attached to the ceiling by a frictionless hinge, so the rod is free to swing back and forth. Pendulum B is a simple pendulum whose length is also d. Obtain the ratio \(T_{A} / T_{B}\) of their periods for small-angle oscillations.

Short Answer

Expert verified
The ratio of their periods \( T_A/T_B \) is \( \sqrt{\frac{2}{3}} \).

Step by step solution

01

Expression for the Period of a Physical Pendulum

The period of a physical pendulum is given by the formula \( T_A = 2\pi \sqrt{\frac{I}{mgL}} \), where \( I \) is the moment of inertia, \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( L \) is the distance from the pivot to the center of mass. For a uniform rod pivoted at one end, the moment of inertia \( I = \frac{1}{3}md^2 \) and \( L = \frac{d}{2} \). Substituting these into the expression gives \( T_A = 2\pi \sqrt{\frac{\frac{1}{3}md^2}{mg \cdot \frac{d}{2}}} \).
02

Simplify the Expression for \(T_A\)

By simplifying the expression for \( T_A \), you get:\[ T_A = 2\pi \sqrt{\frac{\frac{1}{3}md}{\frac{mg}{2}}} = 2\pi \sqrt{\frac{2d}{3g}} = 2\pi \cdot \sqrt{\frac{2}{3}} \cdot \sqrt{\frac{d}{g}}. \]
03

Expression for the Period of a Simple Pendulum

The period of a simple pendulum is given by \( T_B = 2\pi \sqrt{\frac{d}{g}} \). This formula assumes small-angle oscillations where simplicity allows us to ignore complexities of motion.
04

Calculate the Ratio \(T_A / T_B\)

Divide the period of the physical pendulum by the period of the simple pendulum to find the ratio:\[ \frac{T_A}{T_B} = \frac{2\pi \cdot \sqrt{\frac{2}{3}} \cdot \sqrt{\frac{d}{g}}}{2\pi \cdot \sqrt{\frac{d}{g}}} = \sqrt{\frac{2}{3}}. \]
05

Conclusion

The ratio of the periods of the physical pendulum and the simple pendulum is \( \sqrt{\frac{2}{3}} \). This ratio shows how the distribution of mass affects the property's displacement characteristics in harmonic motion.

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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 consists of a mass, often called a 'bob,' suspended from a string or rod of fixed length, with the mass assumed to be a point particle. It is one of the most basic forms of oscillatory systems. As the pendulum swings back and forth, it exhibits a type of periodic motion, characterized by a specific period.The period of a simple pendulum is determined by the formula:\[ T = 2\pi \sqrt{\frac{L}{g}} \]Here, \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity. This formula reflects that the period depends only on the length of the pendulum and the gravitational pull, assuming small deviations from the vertical.
  • The motion is simple harmonic when the angle of displacement is small.
  • This independence from mass makes it a very predictable and straightforward system.
These characteristics make the simple pendulum an ideal model system for studying harmonic motion in physics.
Physical Pendulum
A physical pendulum is a more general form of a pendulum where the mass is distributed along its length, rather than concentrated at a single point. This makes the calculation of its period more complex. In the case of a uniform rod pivoted at one end, it is a classic example of a physical pendulum.To calculate the period of a physical pendulum, we use:\[ T = 2\pi \sqrt{\frac{I}{mgL}} \]Where:
  • \( I \) is the moment of inertia of the pendulum about the pivot point.
  • \( m \) is the total mass.
  • \( g \) is the acceleration due to gravity.
  • \( L \) is the distance from the pivot to the center of mass.
Because of its dependence on the distribution of mass, the period can be different from that of a simple pendulum even when the lengths are the same. For a homogeneous rod, the moment of inertia is calculated differently than for a point mass, affecting the pendulum's behavior under gravitational influence.
Moment of Inertia
The moment of inertia is a critical concept in understanding physical pendulums. It is a measure of an object's resistance to changes in its rotation. In essence, it quantifies how the mass is distributed with respect to a pivot point.For a rod of uniform density and length \( d \) pivoted at one end, the moment of inertia is calculated by:\[ I = \frac{1}{3}md^2 \]
  • Here, \( m \) is the mass of the rod.
  • \( d \) is the length of the rod.
The larger the moment of inertia, the slower an object will spin if the same force is applied. This affects the period of oscillation of the pendulum, as seen in the expression for the period of a physical pendulum. The moment of inertia must be carefully computed to accurately predict the pendulum's motion under oscillation.
Small-Angle Oscillations
Small-angle oscillations are a simplifying assumption used in analyzing pendulums — both simple and physical. It means that the angle of swing is small enough that the sine of the angle is approximately equal to the angle itself when measured in radians.This assumption allows the motion of the pendulum to be treated as simple harmonic motion, which significantly simplifies the equations involved. For small angles (typically less than about 15 degrees), the formulas used to calculate the period of the pendulum accurately predict the behavior of the system under oscillation.
  • For a simple pendulum, this leads to: \( T = 2\pi \sqrt{\frac{d}{g}} \).
  • For a physical pendulum, the small-angle approximation helps in using \( T = 2\pi \sqrt{\frac{I}{mgL}} \).
In practical terms, maintaining small-angle oscillations ensures that the motion stays harmonically simple and calculations remain straightforward.

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Most popular questions from this chapter

\(\mathrm{A} 1.00 \times 10^{-2}-\mathrm{kg}\) block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 124 \(\mathrm{N} / \mathrm{m}\) . The block is shoved parallel to the spring axis and is given an initial speed of 8.00 \(\mathrm{m} / \mathrm{s}\) , while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

In a room that is 2.44 m high, a spring (unstrained length \(=0.30 \mathrm{m} )\) hangs from the ceiling. A board whose length is 1.98 \(\mathrm{m}\) is attached to the free end of the spring. The board hangs straight down, so that its \(1.98-\mathrm{m}\) length is perpendicular to the floor. The weight of the board (104 \(\mathrm{N} )\) stretches the spring so that the lower end of the board just extends to, but does not touch, the floor. What is the spring constant of the spring?

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is \(1.25 \mathrm{m} / \mathrm{s},\) and its maximum acceleration is 6.89 \(\mathrm{m} / \mathrm{s}^{2}\) . How much time elapses between an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum?

When an object of mass \(m_{1}\) is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequency of 12.0 \(\mathrm{Hz}\) . When another object of mass \(m_{2}\) is hung on the spring along with the first object, the frequency of the motion is 4.00 \(\mathrm{Hz}\) . Find the ratio \(m_{2} / m_{1}\) of the mases.

\(\mathrm{A} 1.00 \times 10^{-2}\) -kg bullet is fired horizontally into a \(2.50-\mathrm{kg}\) wooden block attached to one end of a massless horizontal spring \((k=845 \mathrm{N} / \mathrm{m})\) The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.200 \(\mathrm{m}\) . What is the speed of the bullet?

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