CT 2.1.1
You get the strobe timed so an object
appears stationary.
What if we speed up the strobe’s
frequency just a little?
a) The object still staying stationary
b) The object alternating back and forth
between two extremes
c) The object moving forwards
d) The object moving backwards
e) Can’t really guess
Answer: By increasing the strobe’s frequency just a
little, you then see the object a little earlier then
you previously did. Each flash, you see the
object just before it gets to where thing appear
stationary, so the object appears to be going
backwards. (What do you think you would see if
you continued to increase the frequency? )
CT 2.1.2
When the amplitude of an oscillating
object is doubled, the period becomes:
a) twice as big
b) 1/2 as big
c) Stays the same
d) 1/4 times as big
e) Not enough information to decide
Answer: Changing the amplitude of a wave does
nothing to its period.
When
the frequency is doubled,
the period becomes…
Answer: The relationship between frequency and
period is
f=1/T
Doubling f means that T must be cut in half
CT 2.1.4
CT 2.1.3
If I lower the frequency of this tone what
happens to the pitch?
a)
b)
c)
d)
It decreases
It increases
It remains the same
Not enough information
Answer: Frequency is predominately what
determines pitch, so lowering the frequency also
lowers the pitch
What if I make the same sound louder
(keeping the same pitch),
the frequency…
Answer: making the sound louder (or increasing the
amplitude) does nothing to the frequency or pitch
A vibration has a frequency of 100
cycles/second, what is the period (or the
amount of time for one cycle)?
a) 100 sec
b) 1 sec
c) 10 msec
d) 100 msec
e) None of these
Answer: The relationship between frequency and
period is:
f=1/T

T=1/f
= 1 / 100 Hz
= 0.01 sec = 10 msec
(Can you do this conversion? If not ask or find
out how)
Is this tone audible? If so, is it high or low?
Answer: 100 Hz is low frequency, but easily audible
100 Hz is in a base singer’s range
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CT 2.1.5
You are standing outside playing a recorder*
at a steady, perfect “concert A” (that’s 440
Hz, i.e. during one second 440
compressions of air move outwards).
What is the approximate wavelength of the
sound wave produced?
(Data: the speed of sound is 344 m/s)
a) a few mm
b) about half a km
c) A little over one meter
d) A little under one meter
e) None of the above
Answer: We expect that the wavelength will be
about the same size scale as the instrument.
For an exact answer:
v=λf

λ= v/ f
= (344 m/s) / (440 Hz)
= 0.78 m
CT 2.1.6
Speed = wavelength*frequency
v=λf
The air warms up by 20o C. It turns out that
this instrument (like most wind instruments)
produces sound waves of a particular
wavelength.
So, changing the temperature will not
change the wavelength of sound waves
produced by this instrument (noticeably.)
What will change?
a) Frequency
b) Speed
c) Both
Answer:
Air temperature changes sound speed. If v
allowed to change, but λ is not, then f must
d)isNeither
too in order to keep the equation
e)change
??
balanced. It’s not ALWAYS true that λ is fixed
(you have to think about the physics of the
instrument, we’ll talk more about this!), so this
answer is not universal.
CT 2.1.7
We know that pressure changes (as time
goes by) for sound waves.
What changes (as time goes by) for gentle
water waves on a pond?
a) The water density
b) The water temperature
c) The water level or height
d) The water molecules’ position in the
direction of wave propagation
e) Nothing about water waves vary with time
Answer: A water wave is a transverse wave. The
water on the surface moves up and down as the
wave moves across.
CT 2.1.8
Looking at the following waveform, what
is the period?
Amplitude
1
2 time (sec)
a) 1 sec
b) 2 sec
c) 1 m/s
d) 2 m/s
e)Not enough information
Answer: The period is the amount of time that it
takes for the wave to make 1 complete cycle.
This takes 2 sec. in the picture
2
CT 2.1.9
CT 2.1.10
Looking at that same wave (shown again
below), what is its speed?
The wavelength, λ, is 10 m. What is the
speed of this wave?
Amplitude
Amplitude
1
a)
b)
c)
d)
e)
2
Time (sec)
Time (sec)
1/2 m/s
2 m/s
5 m/s
20 m/s
Not enough information
Answer: we need to know the wavelength too
Now given that λ=10 m what is the speed of
the wave?
Answer: The wave speed is:
v = λ.f = (10 m).(1 cycle / 2 sec) = 5 m/s
1
a)
b)
c)
d)
e)
1 m/s
7 m/s
10 m/s
15 m/s
None of the above/not enough info?
Answer: The wave repeats itself 1.5 times in 1
second, so the frequency is:
f = (1.5 cycles) / (1 second) = 1.5 Hz
The wave speed is then:
v = λ.f = (10 m).(1.5 Hz) = 15 m/s
CT 2.2.1
a)
b)
c)
d)
e)
Amplitude vs. Position
Wavelength vs. Time
Voltage vs. Position
Voltage vs. Time
Wavelength vs Amplitude
Answer: An oscilloscope measure and plots the
voltage at as a function of time. Also, if you
connect a microphone to the scope, then the
voltage is proportional to the air pressure (also
referred to as the amplitude of the sound wave..)
What is the difference between the pink
and green lines?
Amplitude
An oscilloscope presents a graphical
representation of:
CT 2.2.2
You do not need to know this, but if you are
curious…
time
a)
b)
c)
d)
e)
Frequency
Amplitude
Period
Phase
Something else
Answer: Both curves have the same amplitude,
wavelength frequency and period. They are just
out of phase with one another.
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CT 2.2.3
What is the period of this wave?
CT 2.2.4
Amp
t=0
What is the period?
Amp
t2
t1
time
0
time
t1
a)
b)
c)
d)
e)
t1
t2
Not at all defined
Not well defined, but t1 is the best
answer
Not well defined, but t2 is the best
answer
Answer: This is not a periodic wave, so technically the
period is not well defined, but because the wave
damps just a little, t1 is a reasonable value for a
period. Also, the longer that t2 gets compared to t1,
the more this will look periodic, and the more well
defined the answer is (and, it’s t1!)
a)
b)
c)
d)
e)
t2
t3
t1
t2
t2-t1
t3-t1
None of the above
Answer: The wave is exactly the same at t1 and t3,
so the difference in these times must be the
period. The wave “repeats” everything that it
did between t1 and t3.
CT 2.4.1
Which of the following is necessary to make
an object oscillate?
i. A stable equilibrium
ii. Little or no friction
iii. A disturbance
a)
b)
c)
d)
e)
i. only
ii. Only
iii. Only
i and iii only
all three
Answer: There must be an equilibrium which the
system tries to approach, also, having no friction
leaves no place for energy to be lost or radiated,
and a disturbance sets the system in motion
Given the above, will the motion be “Simple Harmonic
Motion”?
Answer: The motion may not necessarily be Simple
Harmonic Motion even if the three criteria from
above are satisfied.
CT 2.4.2
The mass and spring shown below are in
equilibrium, a brief downward force is applied
to the mass to put the object into oscillation,
in what direction is the restoring force?
equilibrium
a)
b)
c)
d)
e)
m
Down
Up
Zero
Some other direction
Not enough information
m
moving
down
Answer: pulling the mass down below its equilibrium
makes the spring pull up in an attempt to restore
equilibrium
4
CT 2.4.3
Looking at the same spring and mass
scenario; the spring is stretched and
released. After a moment, when the mass
is on its way up, what is the restoring force?
equilibrium
a)
b)
c)
d)
e)
Down
Up
Right
m
Left
Some other direction
m
moving
up
Answer: The mass is still below equilibrium, so the
spring pulls it up
Answer: When the cart is at E the spring is in
equilibrium so it is neither compressed nor
stretched. The spring holds no potential energy.
The car, however, is moving to the right, so all
of its energy is kinetic energy.
Answer: When the cart is at M it is instantaneously
stopped and the spring is fully compressed.
The car and spring are not moving, so there is
no kinetic energy and since the spring is
completely compressed, all of the energy is
stored as potential energy in the spring.
Answer: When the string is maximally stretched, it is
instantaneously at rest, so it has no kinetic
energy. Since the string is stretched and not
moving it contains only potential energy.
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CT 2.4.4
You are going skydiving. While you are free
falling what is happening to the energy in the
system?
a)
b)
c)
d)
e)
Answer: At the instant that the string is flat, it is
moving through its equilibrium position. There
is no potential energy in the spring because it is
at equilibrium, (not stretched, no stored or
potential energy!) but because it is moving, all
the energy is purely kinetic, which is “energy of
motion”.
Potential energy is being transferred to
kinetic energy
Kinetic energy is being transferred to
Potential energy
The total energy of the system is decreasing
The total energy of the system is increasing
??
Answer: When you are in the plane and not falling all of
your energy is potential energy since you are so
high. (We are neglecting the SIDEWAYS motion you do have some kinetic energy because of that.
Let’s imagine we’re jumping out of a stationary
balloon so we can ignore that!) Once you jump and
begin losing altitude, you pick up speed and gain
kinetic energy. Conservation of energy tells us that
no energy is ever lost or gained, all that’s happening
is a steady conversion of (gravitational) potential
energy into kinetic energy. If air drag gets big, you
are instead converting the potential energy into
THERMAL energy, as you stop speeding up.
CT 2.4.5
Freq. = (constant) * √ ( K/m)
A mass on a spring oscillates with a
certain amplitude and a certain
frequency. If the mass is replaced
with one 4 times heavier, what
happens to the frequency…
A doubles
B: increases by 4
C: halves.
D: decreases by 4
E: None of these/not sure
Answer: Replacing m with 4m gives
Fnew = (const).√ ( K/4m) = (const).(1/2).√ ( K/m) =
(1/2).f
What happens to the period?
Answer: The period is
Tnew =1/ fnew = 1/ [(1/2).f] = 2.(1 / f) = 2 T
CT 2.4.6
F = (constant) * √ ( K/m)
A mass on a spring oscillates with a
certain amplitude and a certain
period T. If the mass is doubled, the
spring constant of the spring is
doubled, and the amplitude of
motion is doubled, the period ..
A: increases
B: decreases
C: stays the same.
D: Not enough information to decide
Answer: doubling K and m together means that
√ ( K/m)  √ ( 2.K/2.m) = √ ( K/m)
no change
cancel the 2s
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