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Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING
Notes: So first half of this HW is still on ideas from oscillations. Problems 2 and 3 blends in
ideas from oscillations with some ideas about collisions from last semester. Problems 4-6 touch upon
basic ideas on waves. For Oscillations it is still Knight (14.1-14.6) or Tipler (14.1-14.3). For waves in
this HW, Knight (20.1-20.3) and Tipler (15.1-15.2).
1. Consider a ball of mass 100 grams tied to a very light string from the ceiling. The length of the
string is 1 meters. (This is just like a simple pendulum setup). I move the ball to the right to an angle 10
degrees and release it (at t=0) , so it starts oscillating (see figure).
(a) What is the speed of the ball 0.2 seconds after I release it?
[This solution contributed by Eric Kuo] I know how to find the position of the ball as a function of
time. From there I can find the velocity at any time. There are two ways to think about this problem:
using forces or using torques. They both lead to the same answer, but in your homework you probably
went one way or other. Feel free to check the method you did on your homework, but before you read
the other solution, I suggest you try it for yourself! I think the solution will do more for you if it’s helping
you on something that you’ve tried yourself.
Thinking about FORCES:
10 degrees is a small enough angle that we can approximate the
motion of the ball as a horizontal oscillation. There are lots of ways to
think about why we can approximate this. Here’s one: solve for the x and y
component of the ball’s position using sin and cos. Then, use the small
angle approximations: sinθ ≈ θ and cos θ ≈1. What is this approximation telling us? That the height of the ball is
almost constant (L) for all small angles, but the horizontal position is
changing as the angle changes (x = L θ). This is why we can approximate this as horizontal oscillation.
So if the ball is only accelerating in the x-direction, then the net force in the y-direction must be
zero, or else the ball would be accelerating up and down. (Ok, so in real life, the
ball is actually moving up and down a little, so there is a net force in the y-
direction. This is really the approximate net force that goes with our
approximation of horizontal oscillation). We know that the y-component of
tension of the string must be equal to mg, since it has to cancel the weight of the
ball out. Using the angle of the string, we can find the x-component of the
tension with a handy triangle.
So the net force is in the x-direction and it’s equal to Fx = –mgθ =
Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING
-mgx/L. Question: why is the force negative? Well, let’s say that the origin is where the angle of
the pendulum is zero. When the position of the ball is to the right, the force on it is to the left and vice
versa; that’s what a restoring force does. Mathematically, this means the sign of x and Fx must be
opposites, so we need a minus sign to make this happen.
[Alternatively, you could think about mgsin(θ) as the force along the direction of the ball’s motion
that is speeding it up as the ball moves. The other forces T-mgcosθ provide the centripetal force;
Using the approximation for small angles the net restoring force redues to mgθ – note that this force changes direction depending on whether the ball is to the right or left of the equilibrium position; The other component, always acts along the direction of tension cannot change the speed of the ball – and so can’t act as the restoring force.]
Now we have the net force, so we do what we do in oscillations: use Newton’s 2nd law to find the
position as a function of time. We know that the answer has to be cos or sin with some constants, so we
just want to figure out what those constants are. We’re thinking about it has a horizontal oscillation, so
we want to write everything in terms of x, and not θ.
The constant A is the amplitude of the oscillation. We know that the amplitude is the value of x
when θ= θ0 = 10 degrees ≈ 0.175 rad ( A = xmax = L θ0). So x(t) = L θ0 cos (√(g/L) t). The velocity of the ball
is v(t) = dx/dt = -(√(gL) θ0 sin (√(g/L) t). To find v(0.2), now just plug in all the numbers in the problem
(g=10 m/s2, L = 1m, t = 0.2 s, θ0 = 0.175 rad). The speed after 0.2 seconds is 0.035 m/s (which is positive, because the negative sign just tells you the direction of the velocity).
Thinking about TORQUES:
The torque on the ball is τ = -rFsinθ = -(L)(mg)sinθ and since 10 degrees is small enough for the
small angle approximation, τ = -Lmgθ. The torque is negative, because it always wants to rotate the
Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING
pendulum back to equilibrium. This means the torque will always have the opposite sign as the angle.
The tension of the string doesn’t cause a torque on the ball because the tension of the string is always
pulling in towards the center of the arc of the pendulum. Torques are applied to objects when the forces
on them cause them to rotate about a pivot point. Only gravity ever has a component in the direction that
the ball is moving. Mathematically, you can see that the tension causes no torque because the angle
between r and Ftension is 180 degrees and sin(180 degrees) = 0.
So we know how much torque is trying to rotate this pendulum. My motivation to find the torque
is that for forces causing oscillations in straight lines we use Newton’s 2nd law, but for rotations we want
to use the rotational version of Newton’s 2nd law, τ = Iα. The value of Iball = mL2 (if you forget why this is, I suggest checking back to your 161 notes on rotational inertia).
Now we do the usual thing that we do in oscillations: we plug everything into Newton’s 2nd law.
We know the answer has to be cos or sin with some constants, so we want to figure out those constants
using the equation we get.
The constant A is the amplitude of the oscillation. We know that the amplitude is the maximum
value of θ. This is when θ = θ0 = 10 degrees ≈ 0.175 rad. So θ (t) = θ0 cos (√(g/L) t). The angular velocity
of the ball is ω(t) = dθ/dt = -(√(g/L) θ0 sin (√(g/L) t). We also know the relationship between velocity and
angular velocity is v = Lω = -(√(gL) θ0 sin (√(g/L) t). Now, just plug in all the numbers in the problem
(g=10 m/s2, L = 1m, t = 0.2 s, θ0 = 0.175 rad). The speed after 0.2 seconds is 0.035 m/s (which is positive, because the negative sign just tells you the direction of the velocity).
NOTES:
I want to point out two things: 1) I did the problem two ways and got the same answer. It’s a good
check not just to see if you’re getting the right answer, but also to see if you really understand the ideas
behind how we are dealing with oscillations and 2) I didn’t plug in all the numbers until the end, and it’s
Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING
not just because I like working with variables (although I do). In the end, I get an expression that tells me
the velocity at any time for a pendulum at any mass with any length and initial displacement.
(b) One student has the following solution: “Well the angular frequency for the pendulum is just ω
= √(g/l) and so the speed of the ball must be v = rω. Here r = l and so v = l√(g/l) = √(gl).” What is wrong
about the student’s reasoning?
Even without going through this student’s reasoning, I can already see that the velocity cannot be
√(gl). When a pendulum swings, it speeds up on the way down and slows down on the way up. This
student is saying that the speed for the pendulum doesn’t depend on time, since g and l are constant.
So where is the mistake in this reasoning? The “ω’s” in ω = √(g/l) and v = rω do not represent the same physical idea. The first is the angular frequency of the pendulum, it is 2π divided by the period of
the pendulum. The second is the instantaneous angular velocity, how fast an object is rotating at some
particular time. If you want to know the speed of the pendulum at a particular time then it should be clear
that you want the second, the instantaneous angular velocity.
In fact, there is no way that the first ω, the angular frequency of the pendulum, can tell you the speed of the pendulum at a specific time. The angular frequency depends only on the period and the
period is just the time it takes to complete one cycle of the pendulum. This isn’t enough information to
tell you how fast the pendulum is moving at a particular time. Here’s an analogy: imagine your
roommate leaves to go to class in the physics building and comes straight back 2 hours later. Even
though you know the time it took for your roommate to leave and come back, you couldn’t say anything
about how fast he/she was walking at a particular time in the trip.
2. (These solutions adapted from Birkett
and Elby) Cart 1, of mass 1.0 kg, rolls down
the ramp shown here (height = 30 centimeters)
and collides with cart 2, of mass 2.0 kg. The
carts stick together and roll toward the spring,
which has spring constant 200 N/m and
equilibrium length 15.0 cm. Neglect friction, air resistance, and the rotational inertia of the wheels (if the
wheels are tiny and light, the rotational inertia can be quite small compared to the inertia of the cart).
(a) Taking rightward as positive, sketch rough, non-numerical graphs of position, velocity, and
acceleration vs. time for cart 2. Start your graph when cart 1 reaches the bottom of the ramp, and
end your graph when the stuck-together carts reach the bottom of the ramp after rebounding off
the spring.
Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING
Before cart 1 reaches cart 2. Cart 2 is just sitting there motionless!
After Collision, but before reaching spring. Steady (constant-velocity) rolling, which means the
position increases steadily, the velocity is constant (flat), and the acceleration is zero. Of course,
there is a spike in acceleration right at the moment of collision. So the cart2 accelerates during the
small time of the collision, after which both carts are travelling at the same speed. Note that during
the collision, we can’t tell for sure the profile of the acceleration, velocity, or position (not enough
information for that) – so I drew something on an average!
Spring compresses. Just like a block on a spring, with the spring compressing from equilibrium.
Spring decompresses. Again, just like a block on spring, starting from rest from its leftmost point.
After leaving the spring, but before reaching ramp. Steady (constant-velocity) rolling in the negative
direction.
(b) Find the closest distance to the wall reached by cart 2.
The carts are closest to the wall at the moment of maximum spring compression, which occurs when the
carts momentarily come to rest while reversing direction. Intuitively, the carts’ speed upon
reaching the spring “controls” how far the spring compresses. Energy conservation bears this out:
As the spring compresses, the kinetic energy carried by the carts converts into elastic potential
energy in the spring. And from that potential energy, we can find the corresponding compression
distance. So, to make progress, we need to know the carts’ speed—and hence, their kinetic
energy—upon first reaching the spring.
Can we find that speed using energy conservation alone? Not quite. Sure, as cart 1 rolls down the
ramp, its gravitational potential energy converts neatly into kinetic energy. But during the inelastic
collision, some of that kinetic converts into internal, thermal energy; the carts heat up slightly.
Since we don’t know how much kinetic energy gets “lost” in this way, we can’t use energy
conservation to find the post-collision speed. But since (i) the force by cart 1 on cart2 is always
equal and opposite to that on cart 2 by cart 1 and (ii) there are no other forces in the horizontal
direction, the change in horizontal momentum of cart 1 is equal and opposite to the change in
horizontal momentum of cart 2. And we can use that to find the speed at which the carts hit the
spring.
In summary, we’ll
(1) use energy conservation to find v1, the speed of cart 1 before the collision;
(2) use momentum conservation to find v2, the speed of the stuck-together carts after the collision; and
(3) use energy conservation to find xmax, the maximum distance by which the spring compresses from
equilibrium. Given xmax, we can figure out cart 2’s closest distance to the wall.
So for the cart 1 that rolls down the ramp, we get
Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING
�
m1 gh =12 m1v1
2 ,
and hence, v1 = 2.42 m/s. This is the velocity of the cart1 at the bottom of the ramp before it hit cart2.
For the collision, since pbefore = pafter, we get m1v1 = (m1 + m2)v2, which we can solve for v2 to get v1/3
= 0.807 m/s. (Hmmm, another way to think: since the combined mass of carts is 3 times the mass of cart 1
alone, after the collision the combined speed is 1/3 the original speed of cart 1)
Now, at maximum compression all the combined kinetic energy of the carts after collision is stored in the
spring as potential energy:
�
12 m1 + m2( )v2
2 =12 kxmax
2 , and hence xmax = 0.099 meters = 9.9 centimeters.
[Alternatively, you could think about the motion of the carts when they are in contact with the springs as
part of a cycle of oscillation – an oscillation where the speed at the equilibrium position is 0.807
m/s. We know that the speed at the equilibrium position is just xmaxω where xmax is the amplitude
of this oscillation and ω is the angular frequency ... and setting that would give us xmax = v2/ω
which if you substitute the angular frequency of that oscillation would turn out to be the same as
you got using energy!]
Finally, since the spring compresses 9.9 cm, cart 2 must be (15-9.9) = 5.1 cm from the wall at the time of
maximum compression.
(c) How much time does cart 2 spend in contact with the spring? Hint: You can think of the time when
the carts are in contact with the spring as part of a period of an oscillation as if the carts were glued
to the spring. Think about what fraction of that period do the carts remain in contact with the
spring.
If the carts became attached to the spring, a complete oscillation would be
�
Equilibirum →1
Leftmost point →2
Equilibirum →3
Rightmost point →4
Equilibirum.
But here, the carts lose contact with spring after completing the second leg of the cycle:
�
Equilibirum →1
Leftmost point →2
Equilibirum.
Our point: The carts stay in contact with the spring for exactly half of a complete oscillation. Therefore,
the time spent in contact with the spring is half a period. So, we need to solve for T and divide it in
half.
Intuitively, the period depends on the stiffness of the spring, since a higher k leads to quicker
oscillations. T should also depend on the mass; common sense says that heavier carts lead to more
lackadaisical, longer-lasting oscillations. When we define angular frequency as
�
ω =km =
200 N/m3.0 kg = 8.16 s -1 ,
Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING
and the period of oscillation is inversely proportional to w.
�
T =2πω
.
Here, the period works out to be 0.77 seconds. The carts stay in contact with the spring for half that time:
t = T/2 = 0.38 seconds.
(d) This “experiment” is repeated, but with weights piled on top of cart 2, to make it heavier.
Compared to your part (c) answer, does cart 2 now spend more time, less time or the same time in
contact with the spring? Answer without doing new calculations.
More time in contact with spring!
As discussed in part (c), the carts stay in contact with the spring for half a period. So, this question asks
whether the new scenario increases or decreases the period of oscillation.
Using equations from above, we can jump to the answer. Since
�
ω = k /m , we get
�
T =2πω
=2πk /m
= 2π mk ,
The spring constant k is the same as before, but the carts’ mass is now bigger. So, the period is now
bigger; the carts spend more time in contact with the spring.
Before moving on, we must see if this answer agrees with common sense, reconciling where necessary.
Does it make sense that increasing the mass of carts makes the oscillations more “lackadaisical”
(longer period)? Absolutely.
[Here is a more subtle question: Does the total distance that the spring compresses change in the case
when cart2 weights more? If it does, would that not also affect the time that the carts stay in
contact with the spring?]
3. (Still working with the same setup as the previous problem – The original setup; NOT the one in
part(d) with added weights) A tenth of a second after cart 2 first comes in contact with the spring,
how far is it from the wall?
While touching the spring, the carts feel a proportional restoring force. Make sure you can work this
force out and with that the expression for the position of the carts when they are in contact with the
spring; don’t just reach for formulas. The relevant conclusion is that, if we set x = 0 at equilibrium,
as drawn here, then the carts’ position while in contact with the spring varies sinusoidally in time,
in accordance with x = Asin wt, x = Acos wt, or some variant of those functions.
Can we use one of those function to solve for x at time t = 0.10 s? Well, the amplitude, A, is the maximum
displacement from equilibrium. We figured it out while solving part (b); the maximum
compression is A = 9.0 centimeters = 0.099 meters. And while solving part (c), we found the
angular frequency. It’s w = 8.16 s–1. So, we can indeed use a function like x = Asin wt or x = Acos wt
to solve for x at a known time. We just need to figure out exactly which of those sinusoidal
functions describes the carts’ motion!
Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING
Let’s set t = 0 when the carts first reach the spring. At that moment, the left end of cart 2 is at x = 0 (the
spring’s equilibrium), and the cart’s position is about to increase. So, the sinusoidal function
describing the carts’ position is the one that starts at 0 and immediately goes up. Only sine fits this
description. So, while in contact with the spring, the carts’ position is given by
�
x = Asinωt .
Using the amplitude and angular frequency discussed above, we get
�
x = Asinωt = (0.099 m) sin 8.16 s-1( ) 0.10 s( )[ ] = 0.072 m = 7.2 centimeters.
That’s not the answer to this question. It’s the distance by which the carts have compressed the spring
leftward from equilibrium – Remember, in that equation we just used, the quantity x is the
displacement from the equilibrium position. The right end of the cart is therefore 15.0 cm – x = 15 cm
– 7.2 cm = 7.8 cm from the wall.
4. [These solutions adapted from Birkett and Elby](Birkett and Elby) Suppose you make waves by
repeatedly dipping your finger into a bathtub of water. Suddenly, you double the rate at which you dip.
(a) Does the velocity of the waves go up, go down, or stay the same? Explain
As we talked about in class, increasing the frequency—i.e., producing more waves per second—does not
make the waves go faster. Wave speed is controlled entirely by the “tightness” (tautness) and
density of the medium through which the waves travel, in this case water. Tapping the water at a
higher frequency does not change the tightness or heaviness of the water. So, the waves travel at
the same speed as before. However. . .
(b) What about the wavelength? Explain.
. . .The waves are now closer together. More precisely, the wavelength gets cut in half. Here’s why,
intuitively. Each dip of the finger creates a new wave-crest (and trough). By doubling the “dip
rate” of the finger, you cut in half the period, i.e., the time between successive dips. Hence, you cut
in half the time that a given that a given crest gets to travel away from the finger before the next
crest is created. Since the crests travel at the same speed no matter quickly or slowly you dip,
cutting in half the time between successive crests also cuts in half the distance between. Which is
another way of saying the wavelength is halved.
The basic wave kinematic equation v = λf when rewritten as λ = v/f, captures this inverse
proportionality between frequency and wavelength; doubling one cuts the other in half. Since v
doesn’t depend on the dip rate, increasing f causes λ to decrease by the same proportion.
(c) Most people have the raw intuition that “faster dipping creates faster motion.” Does your part (a)
answer hopelessly conflict with that intuition, or can you refine that intuition to help you
understand what’s going on? Explain. Hint: It might help some of you to think of a tiny cork bobbing up
and down in the waves.
Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING
The intuition, when refined, is perfectly correct. When you dip at higher frequency, the waves get closer
together, and hence the cork bobs up and down more quickly (higher average speed). So, faster
dipping does create faster motion, where “faster motion” refers to the transverse particle speed (the
individual water molecules bobbing up and down) rather than the wave speed (the speed of
proportional of the pulses along the surface of the water. This is a classic example of refining,
rather than abandoning, our intuitions.
5. (Birkett and Elby) One end of a flexible rope is attached to a wall. By
shaking the other end back and forth at a rate of 3 times per second,
Janice creates the wave pulse shown here. It travels rightward
towards the wall. The gray squares were painted by a mischievous
student on a window-pane immediately behind the rope. Those squares have a side length of 0.10
meters.
(a) What is the amplitude of these waves?
(b) What is the velocity of these waves?
(c) Janice now shakes the rope again, but at a higher frequency. Compared to the waves in part
(a), do these new waves travel faster, slower, or at the same speed?
5. [Solutons contributed by Colleen]
(a) The amplitude of these waves is 0.05 meters. Amplitude is the maximum displacement from the
equilibrium, which can be measured using the squares. The full swing of the wave is one full square
= 0.1m So the deviation from equilibrium is half of that.
(b) Think about a single point on the window pane. 3 wave-crests pass that point per second. So if I
knew the how much distance is there between the crests, I say that he distance travelled by the way
in 1 second equals three times the distance between two successive crests – which gives me the
velocity of the wave. From the snapshot provided, I can tell that each wave is 0.2 meters long
(there are two squares from one crest to another). So the wave’s velocity as it passes is (3 s-1)*(0.2
m) = 0.6 m/s
Or, another way to think about it: We are told that Janice shakes the rope 3 times per second, so it takes
1/3 second for her to create 1 wave pulse. We see that Janice has created 2.5 pulses, which took (2.5
pulses)*(1/3 second per pulse) = 0.83 seconds for the first pulse to get to its current position, which
is 0.5 meters from the start. Therefore, the velocity of the waves is 0.5 meters/0.83 seconds = 0.6
m/s
(c) The waves travel at the same speed! While the frequency is larger, the wavelength is smaller by
exactly the right amount so that the velocity is the same. In general, the speed of a wave through a
medium depends only on the medium’s properties, not on how the wave was produced.
Here is an analogy that could help make sense of it: Think about cars going down the road at some speed.
Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING
Each car here corresponds to a crest on the wave. You could double the number of cars on the
road, still going the same speed, but that means the cars would have to be closer together. This
would double the frequency, or # of cars that pass per minute, without increasing the speed of the
cars.
It is very tempting to keep the shape of the pulse (wavelength) the same and increase the number of
cycles with increase in frequency – and that makes it seem that higher frequency means that the
pulse has gone out further. But if you pay attention to the mechanism of generation of the wave,
you notice that the shape of the pulse (wavelength) would change as Janice shakes the rope more
frequently – generating pulses that have lesser distances between crests, but more number of crests.
It is also tempting to think about how fast a bit of rope is moving. If Janice shakes her hand more
frequently, then the bits of rope must be moving up and down faster (which is correct). And this
sense of fastness (of the bits of rope) can make one think that the wave must have also travelled
faster. But there are two kinds of “faster” involved here: the vertical speed of the bits of rope that
does increases with frequency, and the horizontal speed of the pulse itself which remains constant!
6. (Birkett and Elby) A heavy weight is hanging from a very light string attached to the
ceiling. A worker flicks the string near the ceiling, creating a wave pulse that heads
toward the weight. As the pulse heads downward:
(a) Does its speed increase, decrease, or stay constant? Explain.
[This one I wrote up the solution]
The speed remains the same. The speed of the pulse depends on the tension in the string
and the linear mass density of the string. Neither of which is changing in the case
given. Since the string is very light compared to the block that is hanging, the tension
in the string is relatively unchanged from top to bottom (the top part of the string is
supporting about the same weight as the bottom part). And so the speed of the pulse v = √(T/μ)
where T is the tension and μ the linear mass density of the string, remains unchanged.
(b) Does its amplitude increase, decrease, or stay constant? Explain.
The amplitude of the pulse also remains constant. If I think about two points on a string a little bit apart,
then a change in amplitude of the pulse would mean that the bit of string that is lower goes to a
different maximum distance than the point a little bit above it. If the point at top moves to an
amplitude of 5 cm, there is no way it can push the point below it to move to 6 cm, unless the
tension changes or the bit of string below is lighter. Could the bit below move just 4 cm? If the bit
below moves with an amplitude of 4 cm then something is preventing it from reaching the 5 cm
mark. One option is the pulse loses energy as it goes down so that the point that gets the pulse later
moves through a lower amplitude. But there is nothing in the problem that suggests that the pulse
is losing energy as it moves down. And there is nothing else that would prevent the bit of string
Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING
below to move to the amplitude of 5 cm, or push it further to an amplitude more than 5 cm. In
other words, the amplitude of the pulse does not change!
In reality – well, the rope would have some mass and that would change the situation – How? Well, HW3
solutions would talk more about that. Also, as the pulse passes by parts of the rope are stretched
for some bit of time, before they relax again – and this process heats up the rope by a little bit,
essentially causing the pulse to lose some energy, which would affect the amplitude of the pulse.
But these could be very small effects for a rope of light material and a small amplitude pulse that
does not stretch the rope as much.
(c) A common mistake people make when reasoning about a wave pulse is to think of it (perhaps
unconsciously) as an object, analogous to a ball. How would someone thinking in these object-
based terms answer part (a)? How could you explain why that reasoning is incorrect?
When I drop a ball down, or throw it down it goes faster and faster as it goes down, because the
gravitational force is pulling it down – or it accelerates due to gravity. Someone thinking about the
pulse as an object could reason that gravity would accelerate the pulse, making it move faster as it
goes down.
Alternatively, someone could think about the energy of the pulse. As the ball goes down, the work done
by the gravitational force on the ball increases its kinetic energy (or you could say it loses potential
energy and gains kinetic – it’s all the same really). Similarly, one could argue that as the pulse goes
down, the pulse gains kinetic energy.
The response is that as the pulse is moving down, not a single bit of real matter, stuff, is actually moving
down. Each little bit of the rope moves horizontally, not vertically! So, neither acceleration due to
gravity nor work done by gravitational force come into play here. True – the pulse moves down;
but the pulse is not stuff and the ideas of acceleration as well as kinetic energy from Newton’s Laws
apply to stuff that has mass! It would be weird if someone asked me how much does the pulse
weigh. And if the pulse cannot weight anything, does it make sense to think about its acceleration
due to a gravitational force on it or work done on the pulse by the gravitational force? So, the
reasoning about the pulse gaining speed is incorrect because it attributes mass to the pulse – the
pulse itself is massless, even though bits of stuff do move back and forth as the pulse washes over
them.