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An object suspended from a spring vibrates with simple harmonic motion. At an instant when the displacement of the object is equal to one-half the amplitude, what fraction of the total energy of the system is kinetic and what fraction is potential?

Short Answer

Expert verified
The potential energy is \( \frac{1}{4} \) of the total, and the kinetic energy is \( \frac{3}{4} \).

Step by step solution

01

Understand Simple Harmonic Motion

Simple harmonic motion (SHM) refers to the oscillating motion experienced by the object. The total mechanical energy is conserved and is a sum of kinetic and potential energy as the object oscillates.
02

Define Variables and Equations

Let the amplitude of the oscillation be denoted as \( A \). The displacement \( x \) at the moment in question is \( \frac{A}{2} \). The total mechanical energy \( E \) is constant and can be expressed as \( E = \frac{1}{2}kA^2 \), where \( k \) is the spring constant.
03

Calculate Potential Energy at Displacement \( \frac{A}{2} \)

The potential energy \( U \) in SHM at any displacement \( x \) is \( U = \frac{1}{2}kx^2 \). Substitute \( x = \frac{A}{2} \) into this formula to find potential energy: \[ U = \frac{1}{2}k\left(\frac{A}{2}\right)^2 = \frac{1}{2}k\frac{A^2}{4} = \frac{kA^2}{8}. \]
04

Determine Fraction of Potential Energy

The total energy \( E = \frac{1}{2}kA^2 \). We found \( U = \frac{kA^2}{8} \). The fraction of the total energy that is potential is \( \frac{U}{E} = \frac{kA^2/8}{kA^2/2} = \frac{1}{4} \).
05

Calculate Kinetic Energy

Since the total energy is the sum of potential and kinetic energy, the kinetic energy \( K \) is \( K = E - U = \frac{1}{2}kA^2 - \frac{kA^2}{8} = \frac{4kA^2}{8} - \frac{kA^2}{8} = \frac{3kA^2}{8}. \)
06

Determine Fraction of Kinetic Energy

The fraction of the total energy that is kinetic is \( \frac{K}{E} = \frac{3kA^2/8}{kA^2/2} = \frac{3}{4} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object has due to its motion. In the context of simple harmonic motion (SHM), it is a crucial component of the system's total energy. When an object attached to a spring oscillates, its kinetic energy constantly changes as it moves back and forth.

At the equilibrium position, where the object's displacement is zero, the velocity is at its maximum, meaning all the mechanical energy is kinetic. Conversely, when the displacement is maximum, kinetic energy drops to zero, and potential energy peaks.
  • The formula for kinetic energy is given by: \( K = \frac{1}{2}mv^2 \), where \( m \) is mass, and \( v \) is velocity.
  • In SHM, kinetic energy achieves its highest value at the equilibrium point.
Understanding how to calculate kinetic energy at any point in the oscillation cycle is vital for grasping energy conservation in such systems.
Potential Energy
Potential energy in a spring system represents the energy stored due to the position of an object relative to its equilibrium position. In spring oscillations, this potential energy is specifically referred to as elastic potential energy.

When the spring is compressed or stretched from its neutral position, potential energy is stored. The magnitude of this energy varies with the displacement of the spring from its equilibrium.
  • The potential energy in springs is expressed as: \( U = \frac{1}{2}kx^2 \), where \( k \) is a spring constant and \( x \) is displacement.
  • At maximum displacement, potential energy is at its peak.
It is vital to understand that potential energy transforms into kinetic energy as the spring moves back towards its resting position, showcasing energy conversion dynamics in SHM.
Mechanical Energy Conservation
Mechanical energy conservation is a principle stating that the total energy in an isolated system remains constant if only conservative forces, such as spring forces, act within it.

In SHM involving spring oscillations, the mechanical energy is a combination of kinetic and potential energies.
  • Total mechanical energy \( E \) is expressed as the sum: \( E = K + U \), where \( K \) is kinetic energy and \( U \) is potential energy.
  • This total energy remains constant: \( E = \frac{1}{2}kA^2 \). Given \( A \) is amplitude and \( k \) the spring constant.
This conservation allows us to predict energy distribution in the system, wherein losses in one form of energy are offset by gains in another. Understanding this concept enables us to ascertain energy fractions at various displacement points during oscillation.
Spring Oscillations
Spring oscillations describe the back-and-forth motion experienced by an object attached to a spring. When disturbed from its equilibrium position, the object oscillates in simple harmonic motion due to the restoring force of the spring.

Spring oscillations are characterized by periodic motion, meaning they repeat in a regular cycle.
  • The equation of force in a spring is given by Hooke's Law: \( F = -kx \), enforcing the spring's restoring property.
  • The period and frequency of oscillation are determined by the mass of the object and the spring constant: \( T = 2\pi\sqrt{\frac{m}{k}} \).
Comprehending spring oscillations is key to analyzing and predicting motion dynamics in systems involving buoyancy, acoustics, and even atomic models.

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Most popular questions from this chapter

(a) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern having the same frequency of the note that is sung. If someone sings the note \(\mathrm{B}\) flat that has a frequency of 466 \(\mathrm{Hz}\), how much time does it take the person's vocal cords to vibrate through one complete cycle, and what is the angular frequency of the cords? (b) Hearing. When sound waves strike the eardrum, this membrane vibrates with the same frequency as the sound. The highest pitch that typical humans can hear has a period of \(50.0 \mu \mathrm{s}\). What are the frequency and angular frequency of the vibrating eardrum for this sound? (c) Vision. When light having vibrations with angular frequency ranging from \(2.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) to \(4.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) strikes the retina of the eye, it stimulates the receptor cells there and is perceived as visible light. What are the limits of the period and frequency of this light? (d) Ultrasound. High-frequency sound waves (ultrasound) are used to probe the interior of the body, much as X-rays do. To detect small objects such as tumors, a frequency of around \(5.0 \mathrm{MHz}\) is used. What are the period and angular frequency of the molecular vibrations caused by this pulse of sound?

Compression of human bone. The bulk modulus for bone is 15 GPa. (a) If a diver-in-training is put into a pressurized suit, by how much would the pressure have to be raised (in atmospheres) above atmospheric pressure to compress her bones by \(0.10 \%\) of their original volume? (b) Given that the pressure in the ocean increases by \(1.0 \times 10^{4} \mathrm{~Pa}\) for every meter of depth below the surface, how deep would this diver have to go for her bones to compress by \(0.10 \% ?\) Does it seem that bone compression is a problem she needs to be concerned with when diving?

A \(2.50 \mathrm{~kg}\) rock is attached at the end of a thin, very light rope \(1.45 \mathrm{~m}\) long and is started swinging by releasing it when the rope makes an \(11^{\circ}\) angle with the vertical. You record the observation that it rises only to an angle of \(4.5^{\circ}\) with the vertical after \(10 \frac{1}{2}\) swings. (a) How much energy has this system lost during that time? (b) What happened to the "lost" energy? Explain how it could have been "lost."

A \(100 \mathrm{~kg}\) mass suspended from a wire whose unstretched length is \(4.00 \mathrm{~m}\) is found to stretch the wire by \(6.0 \mathrm{~mm}\). The wire has a uniform crosssectional area of \(0.10 \mathrm{~cm}^{2}\). (a) If the load is pulled down a small additional distance and released, find the frequency at which it vibrates. (b) Compute Young's modulus for the wire.

Find the period, frequency, and angular frequency of (a) the second hand and (b) the minute hand of a wall clock.

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