Physics question thread

[Shrike]

All users may post questions about MCAT and OAT physics here. We will answer the questions as soon as we reasonably can. If you would like to know what physics topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm), though be warned, there are subjects listed there that are rarely tested, or that appear in passages only and need not be learned.

Be sure to check the Physics FAQs and Topic Writeups thread if you have a general question; eventually, many of your answers will be located there. Also, a request: to keep this thread at least somewhat neat, when replying to someone else's post please refrain from quoting anything more than what's necessary for clarity.

Acceptable topics:

• general, MCAT-level physics
• particular MCAT-level physics problems, whether your own or from study material
• what you need to know about physics for the MCAT
• how best to approach to MCAT physics passages
• how best to study MCAT physics
• how best to tackle the MCAT physical sciences section

Unacceptable topics:

• actual MCAT questions or passages, or close paraphrasings thereof
• anything you know to be beyond the scope of the MCAT

Side note: anyone who knows how to post subscripts and superscripts in this system, please PM me the method. I also wouldn't mind knowing how to post some obvious symbols, such as Greek letters and an infinty sign. Should be a matter of changing fonts, but I don't know what's available; again, a PM would be appreciated.

If you really know your physics, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current official contributors to the this thread -- a team to which I hope to add more people:

Thread moderated by: Shrike. Shrike is a full-time instructor for The Princeton Review; he has taken the MCAT twice for no good reason, scoring 14 on the physical sciences section each time. He majored in mathematics, minored in physics, and spent several years accumulating unused school experience (in economics and law).

Also answering questions: Xanthines, a Kaplan instructor. He scored 13 on the PS section of the MCAT and 34 overall.

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[BrokenGlass]

you are very much correct.. this is from kaplan's Qbank, and I just happened to cross this Q last night. The answer IS C (not B).

Do you mean the answer is not 3.8N?

 

[BrokenGlass]

Two objects A and B are placed on a spring, mass A has twice the mass of B. If the spring is depressed and released, object A will:

ExamKrackers states that they will both rise to the same height, but I'm having trouble making this connection. They state that mass is not proportional to height, but won't the elastic potential energy from the spring be converted into GPE (mgh), which will compensate heights due to the masses for each?

Can you post the problem statement and the answer key word for word?

 

[Creightonite]

A question.

When light travels into a denser medium, its velocity decreases (opposite to sound which velocity increases). The textbook says that the wavelength is the one that is changing, not the frequency. Why is that? I could not think of any intuitive answer and the text book does not really explain it.

Also, sound vs. light. Why sound is opposite to light? Sound is based in vibrations so the denser media should have a greater resistance to vibrations. Isually solids are made of heavier molecules than say air and more dense. I was thinking of Young's modulus. The only explanation is that the sound may travel faster in denser media, but it would not travel as far as in a less denser medium -- that's kind of a payback for the increased velocity.

Thank you.

 

[DrChandy]

A question.

When light travels into a denser medium, its velocity decreases (opposite to sound which velocity increases). The textbook says that the wavelength is the one that is changing, not the frequency. Why is that? I could not think of any intuitive answer and the text book does not really explain it.

Also, sound vs. light. Why sound is opposite to light? Sound is based in vibrations so the denser media should have a greater resistance to vibrations. Isually solids are made of heavier molecules than say air and more dense. I was thinking of Young's modulus. The only explanation is that the sound may travel faster in denser media, but it would not travel as far as in a less denser medium -- that's kind of a payback for the increased velocity.

Thank you.

Velocity=Frequency x wavelength

Frequency= velocity/wavelength

Frequency does not change in such a case- its just a matter of how nature behaves.

Using the formula, if velocity increases, wavelength should decrease accordingly to keep frequency constant.
-------------------------
Part B

Again, part of just how nature behaves. Speed of sound is directly proportional to the density of the material (ie speed of sound in steel>speed of sound in air). However, within a phase of a given material (ie solid/liquid/or gas), speed of sound is inversely proportional to density of that material.

Good luck on the MCAT. I took it when it was given in the paper format and eight hours long. Glad to hear theyve changed things.

 

[BrokenGlass]

A question.

When light travels into a denser medium, its velocity decreases (opposite to sound which velocity increases). The textbook says that the wavelength is the one that is changing, not the frequency. Why is that? I could not think of any intuitive answer and the text book does not really explain it.

Also, sound vs. light. Why sound is opposite to light? Sound is based in vibrations so the denser media should have a greater resistance to vibrations. Isually solids are made of heavier molecules than say air and more dense. I was thinking of Young's modulus. The only explanation is that the sound may travel faster in denser media, but it would not travel as far as in a less denser medium -- that's kind of a payback for the increased velocity.

Thank you.

Wavelength of any wave is determined by the medium. Frequency and amplitude of any wave are determined by the wave source.

Speed of any wave is proportional to the square root of the ratio of the elastic component to the inertial component. Sound travels faster in denser media because elastic component is greater in denser media. Here elastic component and inertial component are properties of the medium.

Note: electromagnetic waves don't require a medium in order to travel; in fact, a medium slows EM waves down. EM waves travel the fastest in a vacuum.

 

[xlr8]

Its #67 in the EK physics book if you have that:

Objects A and B are placed on the spring as shown. Object A has twice as much mass as object B. If the spring is depressed and released, propelling the objects into the air, object A will:

A) rise 1/4 as high as object B
B) rise half as high as object B
C) rise to the same height as object B
D) rise twice as high as object B

Answer is C, reasoning:
Take the example to extremes. Here the examples are reasonably close in mass. If object A were 1 million times as massive as objective B, object B will not be propelled 1 million times more into the air than object A, therefore mass is not proportional to height. Since all answers are such, only C can be correct.

I have a problem with that reasoning. Since GPE is involved in the question, which relates mass and height, I try to picture the energy in 2 states: initial and at when both objects are some point in the air. Therefore you'll have GPE (of both masses) + KE (possibly) = Elastic Potential of Spring. From this I think the height one will rise in relation to the other is dependent on mass?

 

[BrokenGlass]

Its #67 in the EK physics book if you have that:

Objects A and B are placed on the spring as shown. Object A has twice as much mass as object B. If the spring is depressed and released, propelling the objects into the air, object A will:

A) rise 1/4 as high as object B
B) rise half as high as object B
C) rise to the same height as object B
D) rise twice as high as object B

Answer is C, reasoning:
Take the example to extremes. Here the examples are reasonably close in mass. If object A were 1 million times as massive as objective B, object B will not be propelled 1 million times more into the air than object A, therefore mass is not proportional to height. Since all answers are such, only C can be correct.

I have a problem with that reasoning. Since GPE is involved in the question, which relates mass and height, I try to picture the energy in 2 states: initial and at when both objects are some point in the air. Therefore you'll have GPE (of both masses) + KE (possibly) = Elastic Potential of Spring. From this I think the height one will rise in relation to the other is dependent on mass?

I don't have the 1001 questions books. What exactly does the picture show?
Is the spring depressed by exactly the same amount in both cases? U = 1/2 * k * x^2, so if x is different, the elastic potential energy will be different for A and B and therefore gravitational potential energy will be different for A and B (since elastic potential energy is converted to gravitational potential energy; I am assuming that objects are launched straight up). So they can rise to the same height. That's certainly possible. What exactly does the picture show?

 

[cheezer]

Its #67 in the EK physics book if you have that:

Objects A and B are placed on the spring as shown. Object A has twice as much mass as object B. If the spring is depressed and released, propelling the objects into the air, object A will:

A) rise 1/4 as high as object B
B) rise half as high as object B
C) rise to the same height as object B
D) rise twice as high as object B

Answer is C, reasoning:
Take the example to extremes. Here the examples are reasonably close in mass. If object A were 1 million times as massive as objective B, object B will not be propelled 1 million times more into the air than object A, therefore mass is not proportional to height. Since all answers are such, only C can be correct.

I have a problem with that reasoning. Since GPE is involved in the question, which relates mass and height, I try to picture the energy in 2 states: initial and at when both objects are some point in the air. Therefore you'll have GPE (of both masses) + KE (possibly) = Elastic Potential of Spring. From this I think the height one will rise in relation to the other is dependent on mass?

Base your reasoning on the equations.

If you call the equilibrium point of the spring y=0, initial velocity=0, assume each mass is propelled from the same spring (read: same spring constant k), and x=the compression of the spring from equilibrium and assume x is different for A and B then:

(1/2)kx^2 + mg(-x) = (1/2)mv^2 + mgh

if A is twice B then:
A = 2m
B = m

Solve the equilibrium equation for h. Plug in 2m, Plug in m. Now you should have two slightly different looking equations solved for h. Set them equal to each other - since h reaches the same height for both - and you'll see that in order for them to reach the same height, mass B will compress the spring to approximately 0.707 the compression of mass A.

Therefore when the masses launch from a spring with the same constant k, they will rise in relation to the compression of the spring.


IF you assume that they are both placed on the spring at the same time and launch, then you plug in 3m instead. You'll see that for the masses to reach the same height, they need to be traveling the same velocity. No way.

Someone feel free to correct me if I'm wrong.

 

[BrokenGlass]

Base your reasoning on the equations.

If you call the equilibrium point of the spring y=0, initial velocity=0, assume each mass is propelled from the same spring (read: same spring constant k), and x=the compression of the spring from equilibrium and assume x is different for A and B then:

(1/2)kx^2 + mg(-x) = (1/2)mv^2 + mgh

if A is twice B then:
A = 2m
B = m

Solve the equilibrium equation for h. Plug in 2m, Plug in m. Now you should have two slightly different looking equations solved for h. Set them equal to each other - since h reaches the same height for both - and you'll see that in order for them to reach the same height, mass B will compress the spring to approximately 0.707 the compression of mass A.

Therefore when the masses launch from a spring with the same constant k, they will rise in relation to the compression of the spring.


IF you assume that they are both placed on the spring at the same time and launch, then you plug in 3m instead. You'll see that for the masses to reach the same height, they need to be traveling the same velocity. No way.

Someone feel free to correct me if I'm wrong.

Well, since energy is a scalar (not a vector), it cannot be negative. So I don't think -x is correct. But your overall approach looks OK. I think you just need to choose a different point as a zero reference point for gravitational potential energy...so the left side of the equation will have only elastic potential energy. Also, my understanding is that we want to know if the final heights are the same for objects A and B, so when they reach their maximum heights, they will no longer have any KE...so the right side should probalby have only graviational potential energy...

but then again, I have been known to be wrong on occasion

 

[cheezer]

Well, since energy is a scalar (not a vector), it cannot be negative. So I don't think -x is correct. But your overall approach looks OK. I think you just need to choose a different point as a zero reference point for gravitational potential energy...so the left side of the equation will have only elastic potential energy. Also, my understanding is that we want to know if the final heights are the same for objects A and B, so when they reach their maximum heights, they will no longer have any KE...so the right side should probalby have only graviational potential energy...

but then again, I have been known to be wrong on occasion

Hmm...
I'm sticking by the -x, because the equation is basically saying that the energy used between the compression and equilibrium of the spring ((1/2)kx^2 + mg(-x)) is converted into KE plus the potential energy of whatever height it obtains.

I also can't think of another reasonable zero reference point, because at equilibrium you can seperate the initial and final mechanical energies to what happens while the mass is on the spring versus what happens when the mass is off the spring.

Also if you where looking at maximum heights, then you'd just get rid of the KE equation on the right side of the equation. In that case, masses will still rise in relation to the compression of the spring. The KE equation only factors in if the masses happen to reach the same height while they're still traveling upwards.

 

[xlr8]

Both masses are on the same spring with a platform (neglect the platform)

|B| | A |
\ \ \ \ \ \ \
/ / / / / / /
\ \ \ \ \ \ \

IF you assume that they are both placed on the spring at the same time and launch, then you plug in 3m instead. You'll see that for the masses to reach the same height, they need to be traveling the same velocity. No way.

That's what I'm saying too, I don't understand why it's B.

 

[cheezer]

Both masses are on the same spring with a platform (neglect the platform)

|B| | A |
\ \ \ \ \ \ \
/ / / / / / /
\ \ \ \ \ \ \



That's what I'm saying too, I don't understand why it's B.

perhaps I spoke too soon about the "no way" part. They start off with the same initial velocity and both are subject to gravity with this equation:

vf^2 = vi^2 -2gy, so basically if height y were the same, the vf is the same and vice versa


blah.

 

[BrokenGlass]

Hmm...
I'm sticking by the -x, because the equation is basically saying that the energy used between the compression and equilibrium of the spring ((1/2)kx^2 + mg(-x)) is converted into KE plus the potential energy of whatever height it obtains.

I also can't think of another reasonable zero reference point, because at equilibrium you can seperate the initial and final mechanical energies to what happens while the mass is on the spring versus what happens when the mass is off the spring.

Also if you where looking at maximum heights, then you'd just get rid of the KE equation on the right side of the equation. In that case, masses will still rise in relation to the compression of the spring. The KE equation only factors in if the masses happen to reach the same height while they're still traveling upwards.

I made a mistake. Potential energy can be negative. One obvious example is U = -GMm/r. Kinetic energy is always >= 0. E = mc^2 is always >=0, but potential energy can be negative...

Regarding this spring question, now that I understand the problem statement I'll post more comments later...

 

[nshams]

Hi everybody! I had a Capacitance/voltage question from Nova Physics - Chapter 15, Passage2, Question#5:

In experiment 1 we place 2 copper circular disks of area A1 a distance d1 apart, thus creating a capacitor with capacitance C1. We connect the two plates to oppostie terminals of a battery which produces a potential Vbat. This produces a positive charge Q1 & an electric field E1 between the plates.

In experiment 3 we reproduce the setup in Experiment 1. THen the wires are removed from the copper plates. We place cellulose nitrate (a dielectric with dielectric constant K=9) between the plates.

What is the magnitude of the potential difference between the plates @ the end of Experiment 3.
A. V1/9
B. 3V1
C. 9V1
D. 81V1

My take: I understand that because the wires are removed in experiment 3 before dielectric is put in, Q3 = Q1. The two equations are:
C1 = Q1*V1

C3 = Q3*V3; but C3 = 9C1 & Q1 = Q3
therefore,: 9C1 = Q1*V3

Divide: 9C1 = Q1*V3 by C1 = Q1*V1

(9C1 / C1) = (Q1 * V3) / (Q1 * V1)
9 = V3 / V1
9V1 = V3

So I said the answer was C but the answer is actually A. WHY?!?!?!?

 

[nshams]

^^ right... so it turns out I had a complete brain fart ... I looked @ the question again & it finally dawned on me that Q = VC NOT C=QV...

this is a bit frightening ... my exam's on May31 :'eek:

 

[rcd]

^^ right... so it turns out I had a complete brain fart ... I looked @ the question again & it finally dawned on me that Q = VC NOT C=QV...

this is a bit frightening ... my exam's on May31 :'eek:

I'm making stupid mistakes too, like thinking C[eq] {and not 1/C[eq]}= 1/C[1] = 1/C[2].

Spent maybe 5 minutes trying to figure how I got a problem testing this wrong .

 

[xlr8]

perhaps I spoke too soon about the "no way" part. They start off with the same initial velocity and both are subject to gravity with this equation:

vf^2 = vi^2 -2gy, so basically if height y were the same, the vf is the same and vice versa


blah.

The kinematics interpretation makes sense, but I still can't see it from the energy perspective. Is this something that should be looked at in terms of kinematics rather than work at first?

 

[rcd]

The kinematics interpretation makes sense, but I still can't see it from the energy perspective. Is this something that should be looked at in terms of kinematics rather than work at first?

for vertical speed v, an object's highest height is independent of its mass.

PE = mgy
KE = .5mv^2

begin with just KE, @highest point with just PE. so conserve energy:

.5mv^2 = mgy

m cancels out.
y = .5 v^2/g.

Since they both start with the same v, and g is the same for everybody on my planet's surface, they both have the same highest height y.

 

[xlr8]

begin with just KE, @highest point with just PE. so conserve energy:

.5mv^2 = mgy

ahhh there you go, since it Uel goes all into KE, it is velocity dependent and mass independent. Thanks to everyone that helped.

 

[ocwaveoc]

Hello
Is that topic not covered in MCAT? I'll gladly forego that topic if we are not responsible.
Thanks!

 

[burtsicle]

Hey guys
So really easy question:
A ball thrown straight up in the air and returns to inital position. You are given the v, t, a (gravity). So in figuring out the total time the ball travels the equation is given as t= 2v/g. What I want to know is where this equation comes from?? How does one deriive it?
Thanx

 

[burtsicle]

Would 3H+ and 3He+ (the 3's are both superscript aka mass number) have the same mass?? If so what is the breakdown of p, n, e??



Thanx

 

[cheezer]

Hey guys
So really easy question:
A ball thrown straight up in the air and returns to inital position. You are given the v, t, a (gravity). So in figuring out the total time the ball travels the equation is given as t= 2v/g. What I want to know is where this equation comes from?? How does one deriive it?
Thanx

yf=yi + vi*t - 0.5gt^2
since the ball returns to its initial position
yf=yi

thus we have 0=vi*t - 0.5gt^2
now calculate
vi*t = 0.5gt^2
vi=0.5gt
2vi/g = t

i think this is correct...

 

[burtsicle]

yf=yi + vi*t - 0.5gt^2
since the ball returns to its initial position
yf=yi

thus we have 0=vi*t - 0.5gt^2
now calculate
vi*t = 0.5gt^2
vi=0.5gt
2vi/g = t

i think this is correct...

cool. you rock. i was doing this :
height = v^2/2g (from KE = PE at max heigth)
then
h=vt (from x=vt)
vt=v^2/2g
t=v/2g
times by 2 for the time required to fall down
t=v/g which is incorrect. I think my initial mistake was assuming h=vt which is only true for horizontal motion.

 

[rcd]

Would 3H+ and 3He+ (the 3's are both superscript aka mass number) have the same mass?? If so what is the breakdown of p, n, e??



Thanx

H is always protons = 1
He is always protons = 2

In neutral species, protons = electrons.
For H+, you have 1 less electron than neutral.
So, H is electrons = 0.
Same thing with He
He electrons = 1.

Neutrons equal mass number - proton numbers.

3H+ = 2
3He = 1

Yeah, their masses would be just about equal, since neutrons have just about the same mass as protons (you also should consider mass defect, but there's not much difference there either).

 

[rcd]

For the optics equation, 1/f = 1/d[o] + 1/d, how do you determine if f is positive or negative?

 

[corbis11]

For the optics equation, 1/f = 1/d[o] + 1/d, how do you determine if f is positive or negative?

converging (convex) lens +f

diverging (concave) lens -f

 

[rcd]

converging (convex) lens +f

diverging (concave) lens -f

Thanks, what about mirrors? Always positive?

 

[BrokenGlass]

Thanks, what about mirrors? Always positive?

converging (concave) mirror +f
diverging (convex) mirror -f

 

[burtsicle]

If a clown is catapulted straight up in air (which is 1.5 m off ground) at 4 m/s. How would you find the velocity at which he hits the ground at (0 meters)??? I was able to figure out his max height as 1.5 + 0.8 =2.3. I also know that it will be more than 4 m/s.
Thanks

 

[Foghorn]

If a clown is catapulted straight up in air (which is 1.5 m off ground) at 4 m/s. How would you find the velocity at which he hits the ground at (0 meters)??? I was able to figure out his max height as 1.5 + 0.8 =2.3. I also know that it will be more than 4 m/s.
Thanks

Use (v final)^2 = v^2 + 2 a y

v = velocity at max height = 0

y = -2.3 (negative since using Cartesian Coordinate w/max ht as ref. pt.)

a = -10 (negative since direction is down) = gravitational acceleration

v final = velocity just before hitting the ground.

 

[rcd]

If a clown is catapulted straight up in air (which is 1.5 m off ground) at 4 m/s. How would you find the velocity at which he hits the ground at (0 meters)??? I was able to figure out his max height as 1.5 + 0.8 =2.3. I also know that it will be more than 4 m/s.
Thanks

Conserve energy. If he was shot up at 4m/s, then when he returns to where he was shot he will be going down at 4m/s, and use gy = .5v^2 to figure out what additional speed he gets in 1.5m.

initial e = final e
PE + KE = PE + KE
mgy + .5mv^2 = mgy + .5mv^2
m cancels out, fill in everything you can
10(1.5) + .5(4)^2 = 0 + .5v^2
15 + 8 = .5v^2
46 = v^2
v < 7m/s.

 

[BrokenGlass]

Hey guys
So really easy question:
A ball thrown straight up in the air and returns to inital position. You are given the v, t, a (gravity). So in figuring out the total time the ball travels the equation is given as t= 2v/g. What I want to know is where this equation comes from?? How does one deriive it?
Thanx

v = v0 + at
v = v0 - gt

0 = v0 - gt (since velocity at max height is zero)
t = v0/g

So now double this time because you want the time for the round trip.

 

[BrokenGlass]

If a clown is catapulted straight up in air (which is 1.5 m off ground) at 4 m/s. How would you find the velocity at which he hits the ground at (0 meters)??? I was able to figure out his max height as 1.5 + 0.8 =2.3. I also know that it will be more than 4 m/s.
Thanks

Chose positive y direction to be down. The magnitude of velocity at 1.5m
is the same whether the clown is going up or down.

v^2 = v0^2 + 2ay (where a is g)
v^2 = 4^2 + 2*10*1.5
V^2 = 46
v is a little less than 7

 

[BrokenGlass]

ahhh there you go, since it Uel goes all into KE, it is velocity dependent and mass independent. Thanks to everyone that helped.

The velocity of the masses at separation is the same because they are launched from the same platform.
Mv^2/2=Mgh, so you can see that M cancels out. The height is independent of mass.

 

[xlr8]

Waves:

One end of a string is shaken each second sending a wave with an amplitude of 10 cm toward the other end. The string is 5 meters long, and the wavelength of each wave is 50 cm. How many waves reach the other end of the string in each 10 second interval?

A. 2
B. 5
C. 10
D. 50


Examkrackers said that the frequency that each wave is sent will be equal at the beginning and end of the string (1 Hz), which I agree with, therefore 10 waves are sent in 10 seconds.

But don't you have to take into account how long it takes for each wave sent to travel the 5 meters, meaning that you would have to first find the velocity of each wave, see how long each takes to go down the string, and then determine how many will reach the other side in 10 seconds?

 

[BrokenGlass]

Waves:

One end of a string is shaken each second sending a wave with an amplitude of 10 cm toward the other end. The string is 5 meters long, and the wavelength of each wave is 50 cm. How many waves reach the other end of the string in each 10 second interval?

A. 2
B. 5
C. 10
D. 50


Examkrackers said that the frequency that each wave is sent will be equal at the beginning and end of the string (1 Hz), which I agree with, therefore 10 waves are sent in 10 seconds.

But don't you have to take into account how long it takes for each wave sent to travel the 5 meters, meaning that you would have to first find the velocity of each wave, see how long each takes to go down the string, and then determine how many will reach the other side in 10 seconds?

There are often many ways to solve a given problem. I don't have the complete solution that EK provided for this problem, but here are my 2 cents:

If you are sending one wave per second, then in 10 seconds you are sending 10 waves.

v = f * lambda = 1 * 0.5 = 0.5 m/s (Unit check: m/s = 1/s * m).

velocity = distance/time

time = distance/velocity = (5 meters)/(0.5 meters/second ) = 10 seconds

So it takes one second for a wave pulse to travel the entire length of the string. Therefore, in 10 seconds, 10 wave pulses reach the other end of the string.

At time = 0 sec, wave pulse 1 is generated.
At time = 1 sec, wave pulse 2 is generated.

At time = 1 sec wave pulse 1 reaches the end of the string
At time = 2 sec wave pulse 2 reaches the end of the string
.
.
.
At time = 10 sec wave pulse 10 reaches the end of the string

 

[xlr8]

There are often many ways to solve a given problem. I don't have the complete solution that EK provided for this problem, but here are my 2 cents:

If you are sending one wave per second, then in 10 seconds you are sending 10 waves.

v = f * lambda = 1 * 0.5 = 0.5 m/s (Unit check: m/s = 1/s * m).

velocity = distance/time

time = distance/velocity = (5 meters)/(0.5 meters/second ) = 10 seconds

So it takes one second for a wave pulse to travel the entire length of the string. Therefore, in 10 seconds, 10 wave pulses reach the other end of the string.

At time = 0 sec, wave pulse 1 is generated.
At time = 1 sec, wave pulse 2 is generated.

At time = 1 sec wave pulse 1 reaches the end of the string
At time = 2 sec wave pulse 2 reaches the end of the string
.
.
.
At time = 10 sec wave pulse 10 reaches the end of the string

The velocity calculated is the velocity for each wave, therefore wouldn't that mean that the 10 seconds is the time required for each wave to traverse the length of the string?

 

[scotties123]

if you have the EK physics book ed.6, page 144 question 173.
It says light travels through a lens and hits point B. Point A is to the left of point B. They then ask what could be done to the lens to make the light hit point A. The correct answer choice is making the lens thicker. The wrong answer choices are bringing the lens closer to the light source, bringing the lens closer to the object (A/B), and making the lens thinner. Im having a really hard time understanding why making the lens thicker is correct and why the other answer choices are wrong. I understand that light wants to travel the shortest possible amount of time, but I dont see how that applies here. Thank you.

 

[gridiron]

if you have the EK physics book ed.6, page 144 question 173.
It says light travels through a lens and hits point B. Point A is to the left of point B. They then ask what could be done to the lens to make the light hit point A. The correct answer choice is making the lens thicker. The wrong answer choices are bringing the lens closer to the light source, bringing the lens closer to the object (A/B), and making the lens thinner. Im having a really hard time understanding why making the lens thicker is correct and why the other answer choices are wrong. I understand that light wants to travel the shortest possible amount of time, but I dont see how that applies here. Thank you.

Hey! I don't have the EK book but I can offer an explanation behind the concept at hand. The purpose of a lense is to diffract light by changing its direction upon entering a lense. This is done by the geometric shape of the lense. Suppose you have light incident on a convex lense (converging lense) from the air. What is happening? The light is passing into a more dense medium of glass. This means the velocity of light will slow down because it is passing from a less dense medium (air) into a more dense medium (glass). By the principle of refraction, this means that light will bend toward the normal line of the lense. The normal line of the lense is the "thick" center of the lense by convention. Conversely, as the light exits the lense it is moving from a more dense medium to a less dense medium which means the light will bend away from the normal line of the lense. This means that the thicker the lense, the more light will bend toward the normal--the more it will be refracted. I hope this helps and good .

 

[scotties123]

Hey! I don't have the EK book but I can offer an explanation behind the concept at hand. The purpose of a lense is to diffract light by changing its direction upon entering a lense. This is done by the geometric shape of the lense. Suppose you have light incident on a convex lense (converging lense) from the air. What is happening? The light is passing into a more dense medium of glass. This means the velocity of light will slow down because it is passing from a less dense medium (air) into a more dense medium (glass). By the principle of refraction, this means that light will bend toward the normal line of the lense. The normal line of the lense is the "thick" center of the lense by convention. Conversely, as the light exits the lense it is moving from a more dense medium to a less dense medium which means the light will bend away from the normal line of the lense. This means that the thicker the lense, the more light will bend toward the normal--the more it will be refracted. I hope this helps and good .

i kinda think i got it now, thanks alot.

 

[BrokenGlass]

The velocity calculated is the velocity for each wave, therefore wouldn't that mean that the 10 seconds is the time required for each wave to traverse the length of the string?

When you think of how fast a wave is moving, think of amount of time it takes for each wave pulse to go from point A to point B. If you shake the rope only once, you send a single wave pulse down the rope. If you keep shaking it, you send multiple wavepulses.

http://www.kettering.edu/~drussell/Demos/reflect/reflect.html

http://www.cord.edu/dept/physics/p128/lecture99_33.html

http://electron9.phys.utk.edu/phys135d/modules/m10/waves.htm

 

[xlr8]

When you think of how fast a wave is moving, think of amount of time it takes for each wave pulse to go from point A to point B. If you shake the rope only once, you send a single wave pulse down the rope. If you keep shaking it, you send multiple wavepulses.

http://www.kettering.edu/~drussell/Demos/reflect/reflect.html

http://www.cord.edu/dept/physics/p128/lecture99_33.html

http://electron9.phys.utk.edu/phys135d/modules/m10/waves.htm

The way im thinking of it is like this:

A pulse is generated. The velocity that we calculated from f*lamba is a measure of the speed of the pulse traveling from where it originated to the end of the string. From that we see that it takes 10 seconds from origination to ending, so the first pulse will reach the end at 10 sec, the next at 11 sec, etc.

 

[BrokenGlass]

The way im thinking of it is like this:

A pulse is generated. The velocity that we calculated from f*lamba is a measure of the speed of the pulse traveling from where it originated to the end of the string. From that we see that it takes 10 seconds from origination to ending, so the first pulse will reach the end at 10 sec, the next at 11 sec, etc.

What's the answer key provided by EK?

 

[xlr8]

C. 10

In the first interval, won't the answer be 1? Since it takes 10 s for the pulse to arrive?

 

[BrokenGlass]

C. 10

In the first interval, won't the answer be 1? Since it takes 10 s for the pulse to arrive?

Lets call the left end of the rope point A and the right end of the rope point B. It takes 10 seconds for each pulse to go from point A to point B

@ t = 0 pulse 1 is generated at point A
@ t = 1 pulse 2 is generated at point A
@ t = 2 pulse 3 is generated at point A
@ t = 3 pulse 4 is generated at point A
@ t = 4 pulse 5 is generated at point A
@ t = 5 pulse 6 is generated at point A
@ t = 6 pulse 7 is generated at point A
@ t = 7 pulse 8 is generated at point A
@ t = 8 pulse 9 is generated at point A
@ t = 9 pulse 10 is generated at point A
@ t = 10 pulse 11 is generated at point A, pulse 1 reaches point B
@ t = 11 pulse 12 is generated at point A, pulse 2 reaches point B
@ t = 12 pulse 13 is generated at point A, pulse 3 reaches point B
@ t = 13 pulse 14 is generated at point A, pulse 4 reaches point B
@ t = 14 pulse 15 is generated at point A, pulse 5 reaches point B
@ t = 15 pulse 16 is generated at point A, pulse 6 reaches point B
@ t = 16 pulse 17 is generated at point A, pulse 7 reaches point B
@ t = 17 pulse 18 is generated at point A, pulse 8 reaches point B
@ t = 18 pulse 19 is generated at point A, pulse 9 reaches point B
@ t = 19 pulse 20 is generated at point A, pulse 10 reaches point B
@ t = 20 pulse 21 is generated at point A, pulse 11 reaches point B
.
.
.
.
.
.
So I guess the answer to this question depends on interpretation. In the first 10 second interval only one wave pulse goes from point A to point B. In all subsequent 10 second intervals 10 wave pulses go from point A to point B.

There is a term in Computer Science called "full pipeline" and it has to do with parallelism (my undergrad and grad degrees are in Computer Science)...I guess EK wants an answer when the pipeline is full (I am just drawing an analogy here).

Maybe other people can comment on this question in case I am misunderstanding what's going on here...

 

[free_radical]

EK says "in each 10 second interval"; im guessing that means time-independent, in other words, once waves are reaching the end already. in which case, math is unnecessary; if every second a wave is sent out, every second a wave has to arrive.

 

[BrokenGlass]

EK says "in each 10 second interval"; im guessing that means time-independent, in other words, once waves are reaching the end already. in which case, math is unnecessary; if every second a wave is sent out, every second a wave has to arrive.

That's certainly one way to look at it, but xlr8's point was that during the inital 10 seconds, it's not the case that 10 waves arrive at the end of the rope. I think EK should/could be more clear with their wording...

 

[free_radical]

you are right- but in such a case on an exam, would we be right to assume that "each 10 second interval" only makes sense if you have multiple 10 second intervals, in which case, the only time you can have only one correct answer is if those multiple 10 second intervals all occur after the initial 10 seconds?

 

[BrokenGlass]

you are right- but in such a case on an exam, would we be right to assume that "each 10 second interval" only makes sense if you have multiple 10 second intervals, in which case, the only time you can have only one correct answer is if those multiple 10 second intervals all occur after the initial 10 seconds?

I would. But I hope AAMC would be clear enough as to what the question is asking so that we can keep the number of assumptions to a minimum...