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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

A body receives impulses of 24Ns and 35Ns inclined 55 degree to each other. calculate the total impulse
Sukpen Reply
what is matar
Abdulrahman Reply
The uniform boom shown below weighs 500 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom?
Jave Reply
A 11.0-m boom, AB , of a crane lifting a 3000-kg load is shown below. The center of mass of the boom is at its geometric center, and the mass of the boom is 800 kg. For the position shown, calculate tension T in the cable and the force at the axle A .
Jave
what is the S.I unit of coefficient of viscosity
Biam Reply
Derived the formula of Newton's law of universal gravitation Fg=G(M1M2)/R2
Monychol Reply
a non-uniform boom of a crane 15m long, weighs 2800nts, with its center of gravity at 40% of its lenght from the hingr support. the boom is attached to a hinge at the lower end. rhe boom, which mAKES A 60% ANGLE WITH THE HORIZONTAL IS SUPPORTED BY A HORIZONTAL GUY WIRE AT ITS UPPER END. IF A LOAD OF 5000Nts is hung at the upper end of the boom, find the tension in the guywire and the components of the reaction at the hinge.
dangly Reply
what is the centripetal force
Malok Reply
Of?
John
centripetal force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
centripetal force is the force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
I do believe the formula for centripetal force is F=MA or F=m(v^2/r)
John
I mean the formula is Fc= Mass multiplied by square of velocity all over the Radius of the circle
Sampson
Yes
John
The force is equal to the mass times the velocity squared divided by the radius
John
That's the current chapter I'm on in my engineering physics class
John
Centripetal force is a force of attraction which keeps an object round the orbit towards the center of a circle. Mathematically Fc=mv²/r
Adebileje
In Example, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m bel
tan Reply
A steel lift column in a service station is 4 meter long and .2 meter in diameter. Young's modulus for steel is 20 X 1010N/m2.  By how much does the column shrink when a 5000- kg truck is on it?
Andiswa Reply
hi
Abdulrahman
mola mass
Abdulrahman
what exactly is a transverse wave
Dharmee Reply
does newton's first law mean that we don't need gravity to be attracted
Dharmee Reply
no, it just means that a brick isn't gonna move unless something makes it move. if in the air, moves down because of gravity. if on floor, doesn't move unless something has it move, like a hand pushing the brick. first law is that an object will stay at rest or motion unless another force acts upon
Grant
yeah but once gravity has already been exerted .. i am saying that it need not be constantly exerted now according to newtons first law
Dharmee
gravity is constantly being exerted. gravity is the force of attractiveness between two objects. you and another person exert a force on each other but the reason you two don't come together is because earth's effect on both of you is much greater
Grant
maybe the reason we dont come together is our inertia only and not gravity
Dharmee
this is the definition of inertia: a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
Grant
the earth has a much higher affect on us force wise that me and you together on each other, that's why we don't attract, relatively speaking of course
Grant
quite clear explanation but i just want my mind to be open to any theory at all .. its possible that maybe gravity does not exist at all or even the opposite can be true .. i dont want a fixed state of mind thats all
Dharmee
why wouldn't gravity exist? gravity is just the attractive force between two objects, at least to my understanding.
Grant
earth moves in a circular motion so yes it does need a constant force for a circular motion but incase of objects on earth i feel maybe there is no force of attraction towards the centre and its our inertia forcing us to stay at a point as once gravity had acted on the object
Dharmee
why should it exist .. i mean its all an assumption and the evidences are empirical
Dharmee
We have equations to prove it and lies of evidence to support. we orbit because we have a velocity and the sun is pulling us. Gravity is a law, we know it exists.
Grant
yeah sure there are equations but they are based on observations and assumptions
Dharmee
g is obtained by a simple pendulum experiment ...
Dharmee
gravity is tested by dropping a rock...
Grant
and also there were so many newtonian laws proved wrong by einstein . jus saying that its a law doesnt mean it cant be wrong
Dharmee
pendulum is good for showing energy transfer, here is an article on the detection of gravitational waves: ***ligo.org/detections.php
Grant
yeah but g is calculated by pendulum oscillations ..
Dharmee
thats what .. einstein s fabric model explains that force of attraction by sun on earth but i am talking about force of attraction by earth on objects on earth
Dharmee
no... this is how gravity is calculated:F = G*((m sub 1*m sub 2)/r^2)
Grant
gravitational constant is obtained EXPERIMENTALLY
Dharmee
the G part
Dharmee
Calculate the time of one oscillation or the period (T) by dividing the total time by the number of oscillations you counted. Use your calculated (T) along with the exact length of the pendulum (L) in the above formula to find "g." This is your measured value for "g."
Dharmee
G is the universal gravitational constant. F is the gravity
Grant
search up the gravity equation
Grant
yeahh G is obtained experimentally
Dharmee
sure yes
Grant
thats what .. after all its EXPERIMENTALLY calculated so its empirical
Dharmee
yes... so where do we disagree?
Grant
its empirical whixh means it can be proved wrong
Dharmee
so cant just say why wouldnt gravity exists
Dharmee
the constant, sure but extremely unlikely it is wrong. gravity however exists, there are equations and loads of support surrounding the concept. unfortunately I don't have a high enough background in physics but have this discussion with a physicist
Grant
can u suggest a platform where i can?
Dharmee
stack overflow
Grant
stack exchange, physics section***
Grant
its an app?
Dharmee
there is! it is also a website as well
Grant
okayy
Dharmee
nice talking to you
Dharmee
***physics.stackexchange.com/
Grant
likewise :)
Grant
Gravity surely exist
muhammed
hi guys
Diwash
hi
muhammed
what is mathematics
Fadumo
What is the percentage by massof oxygen in Al2(so4)3
Isiguzo Reply
molecular mass of Al2(SO4)3 = (27×2)+3{(32×1)+(16×4)} =54+3(32+64) =54+3×96 =54+288 =342 g/mol molecular mass of Oxygen=12×16 =192 g/mol % of Oxygen= (molecular mass of Oxygen/ molecular mass of the compound)×100% =(192/342)×100% =19200/342% =56.14%
Adebileje
A spring with 50g mass suspended from it,has its length extended by 7.8cm 1.1 determine the spring constant? 1.2 it is observed that the length of the spring decreases by 4.7cm,from its original length, when a toy is place on top of it. what is the mass of the toy?
Silindelo Reply
solution mass = 50g= 0.05kg force= 50 x 10= 500N extension= 7.8cm = 0.078m using the formula Force= Ke K = force/extension 500/.078 = 6410.25N/m
Sampson
1.2 Decrease in length= -4.7cm =-0.047m mass=? acceleration due to gravity= 10 force = K x e force= mass x acceleration m x a = K x e mass = K x e/acceleration = 6410.25 x 0.047/10 = 30.13kg
Sampson
1.1 6.28Nm-¹
Anita
1.2 0.03kg or 30g
Anita
I used g=9.8ms-²
Anita
you should explain how yoy got the answer Anita
Grant
ok
Anita
with the fomular F=mg I got the value for force because now the force acting on the spring is the weight of the object and also you have to convert from grams to kilograms and cm to meter
Anita
so the spring constant K=F/e where F is force and e is extension
Anita
mass=50g=50/1000 kg m=0.05kg extension=7.8 cm=7.8/100 e=0.078 m g=9.8 m/s² 1.1 F=ke k=F/e k=mg/e k=0.05×9.8/0.078 k=0.49/0.078 k=6.28 N/m 1.2 F=6.28e mg=6.28e m=6.28e/g e=4.7 cm =4.7/100 e=0.047 m=6.28×0.047/9.8 m=0.29516/9.8 m=0.0301 kg
Adebileje
In this first example why didn't we use P=P° + ¶hg where ¶ is density
Anita Reply
Density = force applied x area p=fA =p = mga, then a=h therefore substitute =p =mgh
Hlehle
Please correct me
Hlehle
sorry I had a little typo in my question
Anita
Density = m/v (mass/volume) simple as that
Augustine
Hlehle vilakazi how density is equal to force * area and you also wrote p= mgh which is machenical potential energy ? how ?
Manorama
what is wave
Alfred Reply
who can state the third equation of motion
Alfred
wave is a distrubance that travelled in medium from one point to another with carry energy .
Manorama
wave is a periodic disturbance that carries energy from one medium to another..
Augustine
what exactly is a transverse wave then?
Dharmee
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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