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When a mass \(m\) is attached to a vertical spring with a force constant \(k,\) it stretches the spring by the amount \(L\). Calculate (a) the period of this mass and (b) the period of a simple pendulum of length \(L\)

Short Answer

Expert verified
(a) \( T = 2\pi \sqrt{\frac{m}{k}} \) for the spring-mass system. (b) \( T = 2\pi \sqrt{\frac{L}{g}} \) for the pendulum.

Step by step solution

01

Understanding the Spring-Mass System

In a spring-mass system, a mass attached to a vertical spring stretches due to gravity until equilibrium is reached. The equilibrium position is reached when the gravitational force is balanced by the spring force.
02

Finding the Stretch of the Spring

The stretch of the spring, denoted as \( L \), is given by Hooke's Law, where \( mg = kL \). Rearranging this, we find the relationship: \( L = \frac{mg}{k} \). This means that the mass stretches the spring by \( L \) under the weight of the mass.
03

Calculating the Period of the Spring-Mass System

The period \( T \) of the mass-spring system is determined using the formula: \( T = 2\pi \sqrt{\frac{m}{k}} \). This equation can be derived from the understanding of simple harmonic motion, where the period depends on the mass \( m \) and the spring constant \( k \).
04

Understanding the Simple Pendulum System

A simple pendulum consists of a mass attached to the end of a rod or string of length \( L \), which swings under the influence of gravity.
05

Calculating the Period of the Simple Pendulum

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

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Key Concepts

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

Spring-Mass System
The spring-mass system is a type of harmonic oscillator involving a spring and a mass. In this setup, a mass is attached to the end of a vertical spring, and it stretches the spring until it reaches an equilibrium position. This equilibrium happens when the gravitational force pulling the mass downward is perfectly countered by the spring force pulling upward. At equilibrium, the system is at rest, but any displacement from this point will start an oscillatory motion.
This type of system is essential in understanding how forces interact in simple harmonic motion. The oscillations occur because the forces involved try to restore the system back to its equilibrium position after being disturbed. It’s important to note that the mass of the object and the stiffness of the spring (described by the spring constant) influence the properties of the oscillation, such as its frequency and period.
Hooke's Law
Hooke's Law is fundamental in understanding how springs work in a spring-mass system. According to Hooke's Law, the force exerted by a spring is proportional to the amount it is stretched or compressed, mathematically expressed as \( F = kx \), where \( k \) is the spring constant, and \( x \) is the displacement from its original position.
In a vertical setup, the mass attached causes a stretch in the spring due to gravity, which is counteracted by the spring's restoring force. Therefore, the equilibrium position can be determined using the equation \( mg = kL \), where \( m \) is the mass, \( g \) is the gravitational acceleration, and \( L \) is how much the spring stretches. Hooke's Law allows us to relate the stretch to the applied forces, giving insight into how much the spring will stretch for a given mass.
Pendulum Period
A simple pendulum is a classic example of simple harmonic motion and is characterized by a mass, known as a bob, attached to a string or rod of fixed length. It swings back and forth under the influence of gravity.
The formula for the period of a simple pendulum is \( T = 2\pi \sqrt{\frac{L}{g}} \). Here, \( L \) represents the length of the pendulum, and \( g \) is the gravitational acceleration. This equation implies that the period is independent of the mass of the bob, depending entirely on the length of the pendulum and the gravitational force. This characteristic makes the pendulum a useful tool in measuring time intervals, as its period remains constant provided the length and gravity remain unchanged.
Gravitational Force
Gravitational force is a fundamental interaction that causes attraction between masses. On Earth, it gives weight to physical objects and governs the motion of celestial bodies. In the context of simple harmonic motion, gravitational force plays a crucial role in systems like the spring-mass and the simple pendulum.
For a spring-mass system, the gravitational force determines the amount the spring stretches to reach equilibrium, given by the formula \( F = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity. Similarly, in a pendulum, gravity acts as the restoring force that causes oscillation. Understanding the gravitational force helps in predicting how these systems will behave under different conditions, such as changes in mass or the strength of the gravitational field.

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Most popular questions from this chapter

An object executing simple harmonic motion has a maximum speed \(v_{\text {max }}\) and a maximum acceleration \(a_{\max }\). Find (a) the amplitude and (b) the period of this motion. Express your answers in terms of \(v_{\max }\) and \(a_{\max }\).

A 0.363-kg mass slides on a frictionless floor with a speed of \(1.24 \mathrm{m} / \mathrm{s} .\) The mass strikes and compresses a spring with a force constant of \(44.5 \mathrm{N} / \mathrm{m}\). (a) How far does the mass travel after contacting the spring before it comes to rest? (b) How long does it take for the spring to stop the mass?

If a mass \(m\) is attached to a given spring, its period of oscillation is \(T\). If two such springs are connected end to end and the same mass \(m\) is attached, (a) is the resulting period of oscillation greater than, less than, or equal to \(T\) ? (b) Choose the best explanation from among the following: I. Connecting two springs together makes the spring stiffer, which means that less time is required for an oscillation. II. The period of oscillation does not depend on the length of a spring, only on its force constant and the mass attached to it. III. The longer spring stretches more easily, and hence takes longer to complete an oscillation.

The maximum speed of a \(3.1-\mathrm{kg}\) mass attached to a spring is \(0.68 \mathrm{m} / \mathrm{s},\) and the maximum force exerted on the mass is \(11 \mathrm{N}\). (a) What is the amplitude of motion for this mass? (b) What is the force constant of the spring? (c) What is the frequency of this system?

A small cart on a \(5.0-\mathrm{m}\) -long air track moves with a speed of \(0.85 \mathrm{m} / \mathrm{s} .\) Bumpers at either end of the track cause the cart to reverse direction and maintain the same speed. Find the period and frequency of this motion.

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