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When the displacement of a mass on a spring is half of the amplitude of its oscillation, what fraction of the mass's energy is kinetic energy?

Short Answer

Expert verified
The fraction of the mass's energy that is kinetic energy when the displacement is half the amplitude is 3/4.

Step by step solution

01

Write the expressions for potential energy and kinetic energy

In a mass-spring system, the potential energy of the mass is given by the formula: \(Potential\; Energy (PE) = \frac{1}{2}kx^2\) where \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position. The kinetic energy of the mass is given by the formula: \(Kinetic\; Energy (KE) = \frac{1}{2}mv^2\) where \(m\) is the mass of the object and \(v\) is the velocity of the object.
02

Apply the conservation of mechanical energy

According to the conservation of mechanical energy, the total mechanical energy of the mass-spring system remains constant throughout the motion: \(Total\; Energy (TE) = KE + PE\) Since we are given that the displacement of the mass is half of its amplitude of oscillation, then \(x = \frac{1}{2}A\), where \(A\) is the amplitude.
03

Find the total energy of the system

To find the total energy of the system, we can consider the mass when it is at its maximum displacement (\(x = A\)) from the equilibrium position. At this point, the velocity of the mass is zero, so the total energy of the system is stored as potential energy: \(TE = PE = \frac{1}{2}kA^2\)
04

Find the potential energy when the displacement is half the amplitude

Since the displacement is half the amplitude, we can plug \(x = \frac{1}{2}A\) into the potential energy formula: \(PE = \frac{1}{2}k(\frac{1}{2}A)^2 = \frac{1}{2}k\frac{1}{4}A^2 = \frac{1}{8}kA^2\)
05

Calculate the kinetic energy when the displacement is half the amplitude

Using the conservation of mechanical energy and substituting the potential energy at half amplitude, we can find the kinetic energy: \(KE = TE - PE = \frac{1}{2}kA^2 - \frac{1}{8}kA^2 = \frac{3}{8}kA^2\)
06

Find the fraction of kinetic energy to the total energy

Now we need to find the fraction of the mass's energy that is kinetic when the displacement is half the amplitude: \(Fraction = \frac{KE}{TE} = \frac{\frac{3}{8}kA^2}{\frac{1}{2}kA^2} = \frac{3}{4}\) The fraction of the mass's energy that is kinetic energy when the displacement is half the amplitude is \(\frac{3}{4}\).

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Key Concepts

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

Conservation of Mechanical Energy
One of the fundamental principles in physics is the conservation of mechanical energy, which simply states that in a closed system with no external forces acting on it, the total mechanical energy remains constant. In the context of an oscillating mass attached to a spring, this means that the energy within the system transforms between kinetic energy—associated with the movement of the mass—and potential energy—associated with the displacement of the mass from its equilibrium position—but the sum of these two types of energy does not change.

As the mass moves through its oscillation, it reaches a point where its speed is at its maximum, and the potential energy is at its minimum. Conversely, when the mass is at its maximum displacement (the amplitude of oscillation), all the energy in the system is stored as potential energy, and the kinetic energy is zero. Understanding the interplay between kinetic and potential energies through the lens of conservation principles provides a powerful tool in solving problems about oscillations.
Kinetic Energy
Kinetic energy represents the energy that an object possesses due to its motion. In a mass-spring system, when the mass is moving, it has kinetic energy given by the formula \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity. The kinetic energy is highest when the mass passes through the equilibrium position, as this is where it has the greatest speed. When deriving the proportion of kinetic energy at a certain displacement, it is essential to consider the conservation of mechanical energy to calculate how the kinetic and potential energy levels vary with respect to the mass's position.
Potential Energy
Potential energy, on the other hand, is the stored energy of position possessed by an object. In a mechanical oscillator like a mass-spring system, the potential energy is related to the displacement of the mass from its equilibrium position. It is described by the equation \(PE = \frac{1}{2}kx^2\), where \(k\) is the spring constant, and \(x\) is the displacement. The potential energy reaches its maximum value when the mass is at the maximum displacement from the equilibrium position—at the amplitude of the oscillation. This energy decreases when the mass moves towards the equilibrium position and becomes zero when the mass is at the equilibrium position itself, having converted all potential energy into kinetic energy.
Mass-Spring System
A mass-spring system is an ideal model used to demonstrate simple harmonic motion, which is a type of periodic motion. The system consists of a mass attached to the end of a spring that can oscillate back and forth around an equilibrium position. The force exerted by the spring on the mass is proportional to the displacement of the mass, according to Hooke's Law (\(F = -kx\)). The negative sign indicates that this force is restoring—it always acts to bring the mass back to the equilibrium position. This simple setup can exemplify various physics principles, including the conservation of mechanical energy, as well as the relationships between displacement, velocity, kinetic energy, and potential energy.
Oscillation Amplitude
The oscillation amplitude is a measure of the maximum displacement of an oscillating object from its equilibrium position. It is a crucial parameter for any oscillating system, as it affects the potential energy stored in the system and the maximum kinetic energy the mass can obtain. In a mass-spring system, the potential energy is at its peak when the displacement is at the amplitude, since the mass is at its furthest point from the equilibrium position. As shown in the example exercise, understanding how the displacement at half of the amplitude affects the potential and kinetic energy ratios is essential in solving for the energy distribution within the system at different points in the oscillation.

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Most popular questions from this chapter

The period of a pendulum is \(0.24 \mathrm{~s}\) on Earth. The period of the same pendulum is found to be 0.48 s on planet \(X\), whose mass is equal to that of Earth. (a) Calculate the gravitational acceleration at the surface of planet X. (b) Find the radius of planet \(X\) in terms of that of Earth.

A hydrogen gas molecule can be thought of as a pair of protons bound together by a spring. If the mass of a proton is \(1.7 \cdot 10^{-27} \mathrm{~kg}\) and the period of oscillation is \(8.0 \cdot 10^{-15} \mathrm{~s}\), what is the effective spring constant for the bond in a hydrogen molecule?

Imagine you are an astronaut who has landed on another planet and wants to determine the free-fall acceleration on that planet. In one of the experiments you decide to conduct, you use a pendulum \(0.500 \mathrm{~m}\) long and find that the period of oscillation for this pendulum is \(1.50 \mathrm{~s}\). What is the acceleration due to gravity on that planet?

You have a linear (following Hooke's Law) spring with an unknown spring constant, a standard mass, and a timer. Explain carefully how you could most practically use these to measure masses in the absence of gravity. Be as quantitative as you can. Regard the mass of the spring as negligible.

A small mass, \(m=50.0 \mathrm{~g}\), is attached to the end of a massless rod that is hanging from the ceiling and is free to swing. The rod has length \(L=1.00 \mathrm{~m} .\) The rod is displaced \(10.0^{\circ}\) from the vertical and released at time \(t=0\). Neglect air resistance. What is the period of the rod's oscillation? Now suppose the entire system is immersed in a fluid with a small damping constant, \(b=0.0100 \mathrm{~kg} / \mathrm{s}\), and the rod is again released from an initial displacement angle of \(10.0^{\circ} .\) What is the time for the amplitude of the oscillation to reduce to \(5.00^{\circ}\) ? Assume that the damping is small. Also note that since the amplitude of the oscillation is small and all the mass of the pendulum is at the end of the rod, the motion of the mass can be treated as strictly linear, and you can use the substitution \(R \theta(t)=x(t)\), where \(R=1.0 \mathrm{~m}\) is the length of the pendulum rod.

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