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Chapter 11: Problem 44

Show that the natural period of vertical oscillation of a mass hung on a Hookean spring is the same as the period of a simple pendulum whose length is equal to the elongation the mass causes when hung on the spring.

Short Answer

Expert verified
The periods of the vertical spring oscillation and the equivalent pendulum are the same: \( 2\pi\sqrt{\frac{m}{k}} \).

Step by step solution

01

Understanding Hookean Spring

The force exerted by a Hookean spring is given by Hooke's Law: \( F = -kx \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. For vertical oscillations, the weight of the mass \( m \) causing elongation is \( mg \). At equilibrium, \( mg = kx \), so \( x = \frac{mg}{k} \).
02

Finding Spring Oscillation Period

The period \( T_s \) of the mass-spring system is given by the formula \( T_s = 2\pi\sqrt{\frac{m}{k}} \). This formula arises from the general formula for the period of harmonic motion: \( T = 2\pi\sqrt{\frac{m}{k}} \).
03

Finding Period of Simple Pendulum

For a simple pendulum of length \( L \), the period \( T_p \) is given by \( T_p = 2\pi\sqrt{\frac{L}{g}} \). In this exercise, the length \( L \) is equivalent to the elongation caused by the mass on the spring: \( L = \frac{mg}{k} \).
04

Comparing the Two Periods

Substituting \( L = \frac{mg}{k} \) in the simple pendulum period formula, we get \( T_p = 2\pi\sqrt{\frac{\frac{mg}{k}}{g}} = 2\pi\sqrt{\frac{m}{k}} \). This is identical to the period \( T_s \) of the mass-spring system.

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Key Concepts

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

Harmonic Motion
Harmonic motion is a fundamental concept in physics often encountered in various contexts, including springs and pendulums. It refers to the repeated back and forth movement of an object in a regular pattern. The motion can be characterized by its amplitude, frequency, and period.

Here are some basics of harmonic motion:
  • **Amplitude**: The maximum extent of vibration or displacement of the oscillating system from its equilibrium position.
  • **Frequency**: How often the cycle repeats itself in a given time period, often measured in hertz (Hz).
  • **Period**: The amount of time it takes to complete one full cycle of the motion. It is usually represented by the symbol \( T \).
In harmonic motion involving a spring, two main forces dominate:- The restoring force from the spring, which tries to pull the mass back to the equilibrium position.- The inertial force of the mass, which resists changes in motion.Together, these forces create the oscillatory motion we observe in a spring-mass system.
Simple Pendulum
A simple pendulum is a device composed of a mass hanging from a fixed point, allowed to swing freely under the influence of gravity. It's one of the classic examples of harmonic motion because its motion is periodic.

A few key points regarding the simple pendulum include:
  • **Length**: The distance from the pivot point to the mass. The length directly influences the pendulum's period.
  • **Period Formula**: The simple pendulum's period \( T_p \) can be calculated using \( T_p = 2\pi\sqrt{\frac{L}{g}} \), where \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.
  • **Assumption of Small Angles**: For the period formula to hold true, the angle of displacement should be small. This ensures that the restoring force is approximately linear.
By equating the length of the pendulum to the elongation caused by a mass on a spring, we can mathematically show that the periods of both systems match under the conditions of small oscillations.
Spring Constant
The spring constant, often represented as \( k \), is a measure of a spring's stiffness. It quantitively defines the relationship between the force exerted on the spring and the displacement caused. According to Hooke's Law, the force exerted by a spring is proportional to the displacement, expressed as \( F = -kx \).

Here's a closer look at the role and implications of the spring constant:
  • **Units**: The spring constant is measured in newtons per meter (N/m). A higher \( k \) means a stiffer spring which requires more force to produce the same displacement as compared to a spring with a lower \( k \).
  • **Application in Oscillations**: The spring constant directly affects the period of oscillation. A stiffer spring results in a shorter period of oscillation, since \( T_s = 2\pi\sqrt{\frac{m}{k}} \).
  • **Role in Equilibrium**: At equilibrium, the spring's force balances the weight of the suspended mass: \( mg = kx \), allowing us to find the equilibrium elongation \( x = \frac{mg}{k} \).
Understanding the spring constant is crucial in solving various problems involving elastic potentials and oscillatory motion.

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Most popular questions from this chapter

A 300 -g object attached to the end of a spring oscillates with an amplitude of \(7.0 \mathrm{~cm}\) and a frequency of \(1.80 \mathrm{~Hz}\). (a) Find its maximum speed and maximum acceleration. (b) What is its speed when it is \(3.0 \mathrm{~cm}\) from its equilibrium position?

A 500 -g object is attached to the end of an initially unstretched vertical spring for which \(k=30 \mathrm{~N} / \mathrm{m}\). The object is then released, so that it falls and stretches the spring. How far will it fall before stopping? [Hint: The \(\mathrm{PE}_{G}\) lost by the falling object must appear as \(\left.\mathrm{PE}_{e} .\right]\)

Compute the acceleration due to gravity at a place where a simple pendulum \(150.3 \mathrm{~cm}\) long swings through \(100.0\) cycles in \(246.7\) \(\mathrm{S} .\) We have $$ T=\frac{246.7 \mathrm{~s}}{100.0}=2.467 \mathrm{~s} $$ Squaring \(T=2 \pi \sqrt{L / g}\) and solving for \(g\) yields $$ g=\frac{4 \pi^{2}}{T^{2}} L=9.749 \mathrm{~m} / \mathrm{s}^{2} $$

A spring undergoes 12 vibrations in \(40 \mathrm{~s}\). Find the period and frequency of the oscillation. $$ T=\frac{\text { Elapsed time }}{\text { Vibrations made }}=\frac{40 \mathrm{~s}}{12}=3.3 \mathrm{~s} \quad f=\frac{\text { Vibrations made }}{\text { Elapsed time }}=\frac{12}{40 \mathrm{~s}}=0.30 \mathrm{~Hz} $$

A cubical block on an air table vibrates horizontally in SHM with an amplitude of \(8.0 \mathrm{~cm}\) and a frequency of \(1.50 \mathrm{~Hz}\). If a smaller block sitting on it is not to slide, what is the minimum value that the coefficient of static friction between the two blocks can have?
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