1. PHYSICS 231 Lecture 33: Oscillations
Remco Zegers
Question hours:Monday 9:15-10:15
Helproom
PHY 231

2. The (loss of) ability to do work: entropy
entropy: S=QR/T R refers to a reversible process
The equation ONLY holds for
a reversible process.
example: Carnot engine:
Hot reservoir: Shot=-Qhot/Thot (entropy is decreased)
Cold reservoir: Scold=Qcold/Tcold
We saw: efficiency for a general engine: e=1-Qcold/Qhot
efficiency for a Carnot engine: e=1-Tcold/Thot
So for a Carnot engine: Tcold/Thot=Qcold/Qhot
and thus: Qhot/Thot=Qcold/Thot
Total change in entropy: Shot+Scold=0
For a Carnot engine, there is no change in entropy
PHY 231

3. The loss of ability to do work: entropy
Now, consider the following irreversible case:
entropy: S=QR/T
This equation only holds
for reversible processes.
T=300 K
conducting
copper
wire
Qtransfer=1200 J
We cut the irreversible
process up into 2 reversible
processes
T=650 K
Shot+Scold=Qhot/Thot+Qcold/Tcold=-1200/ /300=
=+1.6 J/K
The entropy has increased!
PHY 231

4. Hooke’s law Fs=-kx Hooke’s law If there is no friction, the mass
continues to oscillate back and
forth.
If a force is proportional to the displacement x, but
opposite in direction, the resulting motion of the object is
called: simple harmonic oscillation
PHY 231

5. Simple harmonic motion
displacement x A a) b)
time (s) c)
Amplitude (A): maximum distance from
equilibrium (unit: m)
Period (T): Time to complete one full
oscillation (unit: s)
Frequency (f): Number of completed
oscillations per second
(unit: 1/s = 1 Herz [Hz])
f=1/T
PHY 231

6. Simple harmonic motion
displacement x
5cm 2 4 6 8 10
time (s)
-5cm
what is the amplitude of the harmonic oscillation?
what is the period of the harmonic oscillation?
what is the frequency of the harmonic oscillation?
Amplitude: 5cm (0.05 m)
period: time to complete one full oscillation: 4s
frequency: number of oscillations per second=1/T=0.25 s
PHY 231

7. The spring constant k When the object hanging from the spring is not
moving:
Fspring =-Fgravity
-kd =-mg
k = mg/d
k is a constant, so if we
hang twice the amount
of mass from the spring,
d becomes twice larger:
k=(2m)g/(2d)=mg/d
PHY 231

8. displacement vs acceleration
displacement x A
time (s) -A
Newton’s second law: F=ma  -kx=ma  a=-kx/m
acceleration(a)
kA/m
-kA/m
PHY 231

9. example A mass of 1 kg is hung from a spring. The spring stretches
by 0.5 m. Next, the spring is placed horizontally and fixed
on one side to the wall. The same mass is attached and the
spring stretched by 0.2 m and then released. What is
the acceleration upon release?
PHY 231

10. energy and velocity Ekin(½mv2) Epot,spring(½kx2) Sum 0 ½kA2 ½kA2
½mv ½mv2
½k(-A) ½kA2 A -A
conservation of ME: ½m[v(x=0)]2=½kA2 so v(x=0)=±A(k/m)
PHY 231

11. velocity more general Total ME at any displacement x: ½mv2+½kx2
Total ME at max. displacement A: ½kA2
Conservation of ME: ½kA2=½mv2+½kx2
So: v=±[(A2-x2)k/m]
position X
velocity V
acceleration a +A
-kA/m
±A(k/m) -A
kA/m
PHY 231

12. A x time (s) -A demo: cart on track velocity v kA/m a -kA/m A(k/m)
PHY 231

13. Generally: also add gravitational PE
ME = KE PEspring + PEgravity
= ½mv ½kx mgh
PHY 231

14. An example A 0.4 kg object, connected to a light spring with a spring
constant of 19.6 N/m oscillates on a frictionless horizontal
surface. If the spring is compressed by 0.04 and then
released determine: a) the maximum speed of the object
b) the speed of the object when the spring is compressed
by m c) when it is stretched by 0.015m d) for what
value of x does the speed equal one half of the maximum
speed?
PHY 231

15. circular motion & simple harmonic motion
A particle moves in a circular orbit
with angular velocity , corresponding
to a linear velocity v0=r=A 
The horizontal position as a function
of time: x(t)=Acos=Acos(t) (=t)  v0 vx
The horizontal velocity as a function
of time: sin=-vx/v0
vx(t)=-v0sin=-Asin(t)
Time to complete one circle
(I.e. one period T):
T=2A/v0=2A/A=2/
=2/T=2f (f: frequency)
: angular frequency A x
t=0
PHY 231

16. Circular motion and simple harmonic motion
The simple harmonic motion can be described by the
projection of circular motion on the horizontal axis.
xharmonic(t)=Acos(t)
vharmonic(t)=-Asin(t)
where A is the amplitude of the oscillation, and
=2/T=2f, where T is the period of the harmonic
motion and f=1/T the frequency.
PHY 231

17. For the case of a spring ±A(k/m) 1) velocity is maximum if v=±A(k/m)
position X
velocity V
acceleration a +A
-kA/m
±A(k/m) -A
kA/m
1) velocity is maximum if v=±A(k/m)
2) circular motion: vspring(t)=-Asint maximal if vspring=±A
combine 1) & 2) =(k/m)
Acceleration: a(t)=-(kA/m)cos(t)=-2Acos(t)
PHY 231

18. xharmonic(t)=Acos(t)
time (s) -A
=2f=2/T=(k/m)
velocity v
A(k/m)
vharmonic(t)=-Asin(t)
-A(k/m)
kA/m a
aharmonic(t)=-2Acos(t)
-kA/m
PHY 231

19. Example A mass of 0.2 kg is attached to a spring with k=100 N/m.
The spring is stretched over 0.1 m and released.
What is the angular frequency () of the corresponding
circular motion?
What is the period (T) of the harmonic motion?
What is the frequency (f)?
What are the functions for x,v and t of the mass
as a function of time? Make a sketch of these.
PHY 231

20. question An object is attached on the lhs and rhs by a spring with the
same spring constants and oscillating harmonically.
Which of the following is NOT true?
In the central position the velocity is maximal
In the most lhs or rhs position, the magnitude of the
acceleration is largest.
the acceleration is always directed so that it counteracts
the velocity
in the absence of frictional forces, the object will
oscillate forever
the velocity is zero at the most lhs and rhs positions
of the object
PHY 231