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The motion and energy of a mass attached to a horizontal spring, spring constant k, at various points in its motion. In figure (a) the mass is displaced to a position x = A to the right of x =0 and released from rest (v=0.) The spring is stretched. The force on the mass is to the left. The diagram is labeled with one half k A squared. (b) The mass is at x = 0 and moving in the negative x-direction with velocity – v sub max. The spring is relaxed. The Force on the mass is zero. The diagram is labeled with one half m quantity v sub max squared. (c) The mass is at minus A, to the left of x = 0 and is at rest (v =0.) The spring is compressed. The force F is to the right. The diagram is labeled with one half k quantity minus A squared. (d) The mass is at x = 0 and moving in the positive x-direction with velocity plus v sub max. The spring is relaxed. The Force on the mass is zero. The diagram is labeled with one half m v sub max squared. (e) the mass is again at x = A to the right of x =0. The diagram is labeled with one half k A squared.
The transformation of energy in SHM for an object attached to a spring on a frictionless surface. (a) When the mass is at the position x = + A , all the energy is stored as potential energy in the spring U = 1 2 k A 2 . The kinetic energy is equal to zero because the velocity of the mass is zero. (b) As the mass moves toward x = A , the mass crosses the position x = 0 . At this point, the spring is neither extended nor compressed, so the potential energy stored in the spring is zero. At x = 0 , the total energy is all kinetic energy where K = 1 2 m ( v max ) 2 . (c) The mass continues to move until it reaches x = A where the mass stops and starts moving toward x = + A . At the position x = A , the total energy is stored as potential energy in the compressed U = 1 2 k ( A ) 2 and the kinetic energy is zero. (d) As the mass passes through the position x = 0 , the kinetic energy is K = 1 2 m v max 2 and the potential energy stored in the spring is zero. (e) The mass returns to the position x = + A , where K = 0 and U = 1 2 k A 2 .

Consider [link] , which shows the energy at specific points on the periodic motion. While staying constant, the energy oscillates between the kinetic energy of the block and the potential energy stored in the spring:

E Total = U + K = 1 2 k x 2 + 1 2 m v 2 .

The motion of the block on a spring in SHM is defined by the position x ( t ) = A cos ( ω t + ϕ ) with a velocity of v ( t ) = A ω sin ( ω t + ϕ ) . Using these equations, the trigonometric identity cos 2 θ + sin 2 θ = 1 and ω = k m , we can find the total energy of the system:

E Total = 1 2 k A 2 cos 2 ( ω t + ϕ ) + 1 2 m A 2 ω 2 sin 2 ( ω t + ϕ ) = 1 2 k A 2 cos 2 ( ω t + ϕ ) + 1 2 m A 2 ( k m ) sin 2 ( ω t + ϕ ) = 1 2 k A 2 cos 2 ( ω t + ϕ ) + 1 2 k A 2 sin 2 ( ω t + ϕ ) = 1 2 k A 2 ( cos 2 ( ω t + ϕ ) + sin 2 ( ω t + ϕ ) ) = 1 2 k A 2 .

The total energy of the system of a block and a spring is equal to the sum of the potential energy stored in the spring plus the kinetic energy of the block and is proportional to the square of the amplitude E Total = ( 1 / 2 ) k A 2 . The total energy of the system is constant.

A closer look at the energy of the system shows that the kinetic energy oscillates like a sine-squared function, while the potential energy oscillates like a cosine-squared function. However, the total energy for the system is constant and is proportional to the amplitude squared. [link] shows a plot of the potential, kinetic, and total energies of the block and spring system as a function of time. Also plotted are the position and velocity as a function of time. Before time t = 0.0 s, the block is attached to the spring and placed at the equilibrium position. Work is done on the block by applying an external force, pulling it out to a position of x = + A . The system now has potential energy stored in the spring. At time t = 0.00 s, the position of the block is equal to the amplitude, the potential energy stored in the spring is equal to U = 1 2 k A 2 , and the force on the block is maximum and points in the negative x -direction ( F S = k A ) . The velocity and kinetic energy of the block are zero at time t = 0.00 s . At time t = 0.00 s, the block is released from rest.

Graphs of the energy, position, and velocity as functions of time for a mass on a spring. On the left is the graph of energy in Joules (J) versus time in seconds. The vertical axis range is zero to one half k A squared. The horizontal axis range is zero to T. Three curves are shown. The total energy E sub total is shown as a green line. The total energy is a constant at a value of one half k A squared. The kinetic energy K equals one half m v squared is shown as a red curve. K starts at zero energy at t=0, and rises to a maximum value of one half k A squared at time 1/4 T, then decreases to zero at 1/2 T, rises to one half k A squared at 3/4 T, and is zero again at T. Potential energy U equals one half k x squared is shown as a blue curve. U starts at maximum energy of one half k A squared at t=0, decreases to zero at 1/4 T, rises to one half k A squared at 1/2 T, is zero again at 3/4 T and is at the maximum of one half k A squared again at t=T. On the right is a graph of position versus time above a graph of velocity versus time. The position graph has x in meters, ranging from –A to +A, versus time in seconds. The position is at +A and decreasing at t=0, reaches a minimum of –A, then rises to +A. The velocity graph has v in m/s, ranging from minus v sub max to plus v sub max, versus time in seconds. The velocity is zero and decreasing at t=0, and reaches a minimum of minus v sub max at the same time that the position graph is zero. The velocity is zero again when the position is at x=-A, rises to plus v sub max when the position is zero, and v=0 at the end of the graph, where the position Is again maximum.
Graph of the kinetic energy, potential energy, and total energy of a block oscillating on a spring in SHM. Also shown are the graphs of position versus time and velocity versus time. The total energy remains constant, but the energy oscillates between kinetic energy and potential energy. When the kinetic energy is maximum, the potential energy is zero. This occurs when the velocity is maximum and the mass is at the equilibrium position. The potential energy is maximum when the speed is zero. The total energy is the sum of the kinetic energy plus the potential energy and it is constant.

Questions & Answers

find the angle of projection at which the horizontal range is twice the maximum height of a projectile
Akaji Reply
impulse by height fomula
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Hello
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impulse is the integral of a force (F),over the interval for which it act
Agbeyangi
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magnitude and direction
Chris
what is impulse?
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the product of force and time
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How are you all?
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p=mv
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lasitha
angular displacement = l/r
anand
friend ,what is the " l "?
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I have another problem,what is the difference between pure chemistry and applied chemistry ?
lasitha
l stands for linear displacement and l>>r
anand
r stands for radius or position vector of particle
anand
pure chemistry is related to classical and basic concepts of chemistry while applied Chemistry deals with it's application oriented concepts and procedures.
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what is the deference between precision and accuracy?
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I think in general both are same , more accuracy means more precise
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and lessor error
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lasitha
what are the main concepts of applied chemistry ?
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anand,can you give an example for angular displacement ?
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when any particle rotates or revolves about something , an axis or point ,then angle traversed is angular displacement
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like a revolving fan
anand
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lasitha
o need this application in French language
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Why is it that the definition is not there
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what is physics
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the study of matter in relation to energy
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Ampher law
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multiply by 3.6
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a man drops a rock into a wall
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then
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Atomic theory,
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according to John Dalton all the matter was composed of atom and indivisible, indestructible building blocks. while the atom of different elements have different mass and size.
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two parallel wires are 5.80cm apart and carry currents in opposite direction , as shown in figure .find the magnitude and direction of the magnetic field at point p due to two 1.50mm segments of wire that are opposite each other and each 8.00 cm from p.
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solve it
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the formula velocity tell me
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v=d/t
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in electrostatics
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Practice Key Terms 3

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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