1) A 2.15 kg mass is pulled back 40 cm from a spring and released, oscillating with total energy of 24.8 J. 2) At its maximum displacement of 25 cm from equilibrium, the mass' speed is 3.75 m/s. 3) The mass' kinetic and potential energy will be equal when it reaches 12.4 J, or 28.3 cm from equilibrium.

1. Yangfan(Tony) Sun
26559147
Lab Section:LD2
1
Physics 101 Learning Object 1
Energy in a Mass Spring System
Question:A massis connectedaspringoscillator.The massispulledbackfromthe springat a distance
of 40.0 cm, isreleased,andbeginstooscillate.The springconstanthasa value of 310 kg/s2
.
What isthe total amountof energyof thissystem?
Duringitsoscillation,whatwillitstopspeedbe?Assume the massis2.15 kg.
The mass is25 cm awayfrom the spring’sstable point.Whatisthe speedof itat thismoment?
How far fromthe stable positionwill the massbe if itskineticenergyis equal toitspotential
energy?Tothe leftof equilibriumortothe rightof equilibrium?
If the massundergoessimpleharmonicmotionwithanamplitude of 30 cm and is10 cm away
fromthe spring’sstable position,how muchof the total energyisin the formof potential energy?

2. Yangfan(Tony) Sun
26559147
Lab Section:LD2
2
Physics 101 Learning Object 1
Energy in a Mass Spring System
(Solutions)
Question:A massis connectedaspringoscillator.The massispulledbackfromthe springat a distance
of 40.0 cm, isreleased,and beginstooscillate.The springconstanthasa value of 310 kg/s2
.
What isthe total amountof energy inthissystem?
If we take any positionof the massitspath,we can calculate the total energy(E= K + U) because energy
isconserved.The onlypositionclearly statedinthe questionisatitsinitial positionatt = 0. Thismeans
there isno kineticenergy.Allof the energyisinthe formof potential energy.We know
U = ½ kx2
U = ½ (310) (0.400)2
U = 24.8 J.
The total amountof energyinthissystemis24.8 J.
Duringitsoscillation,whatwillitstopspeedbe? Assume the massis2.15 kg.
If the massis movingattop speed,thenall the energyof thissysteminthe formof kineticenergy.Using
the formula:
v = √ (2E/m)
v = √ (2(24.8)/(2.15))
v = 4.80 m/s
The mass is25 cm awayfrom the spring’sstable point.Whatisthe speedof itat thismoment?
A simple use of the total energyequationcanallow ustosolve this problem.
E = U + K
E = ½ kx2
+ ½ mv2
v2
= (2E – kx2
)/m
v = √ [(2E – kx2
)/m]
v = √ [(2(24.8) – (310)(0.25)2
)/2.15]
v = 3.75 m/s
The speedat thismomentis3.75 m/s.

3. Yangfan(Tony) Sun
26559147
Lab Section:LD2
3
(Solutions continued)
How far fromthe stable positionwill the massbe if itskineticenergyis equal toitspotential
energy?Tothe leftof equilibriumortothe rightof equilibrium?
Usinga graph:
(Picture from“PhysicsforScientistsand EngineersAnInteractiveApproach”textbook)
We can see thatthe graph onthe rightdepictsthe kineticenergyandpotential energyasafunctionof
displacement(inthe formx(t) =A cos(ωt+ φ).The graph is an evenfunction,andthusthe onlytime the
kineticenergyandpotentialenergyare equal iswhenitis half the total energy,between–A andthe
equilibriumandbetweenthe equilibriumandA,or24.8/2 = 12.4 J.
K = E
½ kx2
= E
x = √ (2E/k)
x = √ (2(12.4)/310)
x = 0.283 m
The graph showsthat the same amountof kineticenergyisthe same forequal distancestoboththe left
and rightof the equilibrium.Thisalsoappliestopotential energy.
If the massundergoessimpleharmonicmotionwithanamplitude of 30 cm and is10 cm away
fromthe spring’sstable position, how muchof the total energyisinthe formof potential energy?
U = ½ kx2
E = ½ kA2
U/E = (½ kx2
) /(½ kA2
)
U/E = x2
/ A2
U/E = 102
/ 302
U/E = 1/9
So 1/9 or about11% of the total energyisinthe form of potential energywhenthe massis10 cm away
fromthe stable position.