Join TSR now and get all your revision questions answeredSign up now

AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread] Watch

    Offline

    1
    ReputationRep:
    (Original post by NEWT0N)
    Darn but I thought this was intuitive stuff like running through air
    There seem to be two mistakes that people are making:

    (1) Thinking that damping is a change in the time period of the pendulum. Damping changes the amplitude (damping changes the period slightly, but we're not supposed to know that :P The principle feature is that is changes amplitude)

    (2) Concluding that because the force doing the damping is the same for both (correct), that the effect of that damping is the same for both (incorrect). Everyone is forgetting that a force's effect on an object depends entirely on the object's mass.
    Offline

    1
    ReputationRep:
    Aye to everyone who quoted me you were right about the capacitor-charge one, I overthought it haha.
    • Thread Starter
    Offline

    3
    ReputationRep:
    Guys. Bad news I'm afraid

    Name:  ImageUploadedByStudent Room1434032804.520422.jpg
Views: 1179
Size:  122.5 KB


    Posted from TSR Mobile
    Offline

    13
    ReputationRep:
    (Original post by ubisoft)
    Did it mention air resistance in the question? they can't be that harsh
    It's implied.. what else could possibly damp the pendulum?


    (Original post by Masr97)
    You're saying the reason it dampens is because it adds air resistance? That's pretty weak reasoning if you ask me. If they wanted an answer about air resistance they would've made it more obvious.

    They mentioned no dimensions when they were talking about the ring. Theoretically you could have an EXTREMELY small volume ring with a massive density which would mean the extra air resistance is negligible.

    I'm still gonna go with no change on the damping.
    No, the ring decreases damping. You're arguing against the opposite of what I said :I
    Offline

    0
    ReputationRep:
    Is there an unofficial mark scheme??


    Posted from TSR Mobile
    Offline

    4
    ReputationRep:
    (Original post by JizzaStanger)
    NEWTON is not quite right. He's suggesting you add mass to the pendulum while it is oscillating, which is not correct.

    The reason is that the force that causes the damping is equal for both cones. As we know from F=ma (or rate of change of momentum), a force has a larger effect on an object with a lower mass. The reduction in momentum due to the damping force is the same, but the reduction in velocity is smaller for the larger mass (i.e. with the ring), meaning the reduction in kinetic energy (and thus total energy) is lower, and therefore the damping is less with a higher mass.
    OMG this is what I put! I thought I was babbling but hopefully it's the correct answer! for the first part of the question was the wording which arrangment produced the MOST damping? or the least damping?
    Offline

    1
    ReputationRep:
    (Original post by NEWT0N)
    They would be getting away if you think of it in terms of what could have happened.

    People who got the other parts right could have had 2 marks from that question, and the only mark that AQA would ignore would be the y-intercept of your curve so you would essentially get full marks for the question if you got the other parts right

    However, people who got the whole question wrong could have had 0 marks. That's 2 marks less than people who got it right.

    You cannot then equate these two groups of people by just ignoring the whole question. The people who should've had 2 marks should have more points with respect to that question. Not sure how they'll account for this
    You could apply this logic to the absence of any valid question. Eg. Some people would have got full marks on a question asking them to calculate the time period of a pendulum and some wouldn't have, and thus the non-inclusion of a question asking that wasn't fair on the people who would have got all the marks for it.

    On the whole, the people who were going to get this question wrong were also going to get other questions wrong.
    Offline

    1
    ReputationRep:
    (Original post by aprocrastinator)
    OMG this is what I put! I thought I was babbling but hopefully it's the correct answer! for the first part of the question was the wording which arrangment produced the MOST damping? or the least damping?
    Can't remember. It was worded in a confusing way :P I think it was which has the least damping.
    Offline

    14
    ReputationRep:
    The MEAN emf induced could not have been 0.12V and this is my explanation as to why. The rate of change of flux on the data sheet is given by delta (change in) flux over change in time. This is not a derivative. Hence this gives the emf when the change in flux is considered constant. However, the flux change was not constant as it is a coil rotating; the flux varies with a cosine function as the coil was initially perpendicular to the field. The change in flux linkage was 6x10^-2 Wbturns iirc and the time that this flux changed to 0 in was 0.5s, this gives 0.12V which is as if a conductor was dragged through a field at CONSTANT VELOCITY. However, you can model the coil as two conductors moving in circular motion where their velocity perpendicular to the field varies over time. Since the emf varies sinusoidally over time from t=0 to t=0.5s then the average emf must be 0.12v/sqrt(2) it is the root mean squared which gives 0.08V
    Offline

    13
    ReputationRep:
    (Original post by Protoxylic)
    The MEAN emf induced could not have been 0.12V and this is my explanation as to why. The rate of change of flux on the data sheet is given by delta (change in) flux over change in time. This is not a derivative. Hence this gives the emf when the change in flux is considered constant. However, the flux change was not constant as it is a coil rotating; the flux varies with a cosine function as the coil was initially perpendicular to the field. The change in flux linkage was 6x10^-2 Wbturns iirc and the time that this flux changed to 0 in was 0.5s, this gives 0.12V which is as if a conductor was dragged through a field at CONSTANT VELOCITY. However, you can model the coil as two conductors moving in circular motion where their velocity perpendicular to the field varies over time. Since the emf varies sinusoidally over time from t=0 to t=0.5s then the average emf must be 0.12v/sqrt(2) it is the root mean squared.
    The coil did a quarter-turn, right? So it would just be 0.12 V.
    Offline

    13
    ReputationRep:
    (Original post by NEWT0N)
    I don't get the "horizontal line at 0A" part but this is pretty much what I expected, although I did expect them to be more generous for their mistake and give the "y intercept mark" for any positive y intercept that is lower than the one for the 'smaller resistance curve'
    I hope it's any lower one; I just did it from the first square (no particular reason, it was just there lol).
    Offline

    4
    ReputationRep:
    (Original post by JizzaStanger)
    Can't remember. It was worded in a confusing way :P I think it was which has the least damping.
    uhhh ok, I'm sure I put my intended answer at the time lol but I ticked the middle box I believe...?

    Actually I'm pretty sure I was asking for the one with the most damping. Cos I remember I ticked the box which was the longest sentence so it would've been 'cone without the ring' ... dear god this is confusing
    Offline

    14
    ReputationRep:
    (Original post by StarvingAutist)
    The coil did a quarter-turn, right? So it would just be 0.12 V.
    Yes, so a quarter of a sin wave. The EMF varies over that time from 0 to a maximum emf at 0.5s. Prove for yourself that if the flux linkage is ACos(wt) for constant A=BAN. Then EMF=-dN(phi)/dt which is AwSin(wt) the emf is not constant. Hence the need to find the mean emf
    Offline

    3
    ReputationRep:
    (Original post by CD223)
    Guys. Bad news I'm afraid

    Name:  ImageUploadedByStudent Room1434032804.520422.jpg
Views: 1179
Size:  122.5 KB




    Posted from TSR Mobile

    I think that's the best way to solve it tbh
    Offline

    1
    ReputationRep:
    (Original post by NEWT0N)
    Thankfully I stated that the period stays the same so hopefully a mark there.

    So it boils down to a heavier mass experiencing more damping?

    Also would you mind saying what you think the marks are given for?

    1 mark for ticking the correct box
    1 mark for saying that damping force is constant
    1 mark for saying time period is constant
    and 1 mark for explaining how the constant damping force damps the larger mass less?
    The heavier mass experiences less damping.

    1 mark for ticking the correct box
    1 mark for saying that damping force is same for cone with and without ring.
    1 mark for some kind of application of Newton's second law, OR saying that the damping force does work against the cones, but has less effect on the velocity of the ringed cone because KE is proportional to mass.
    1 mark for saying total energy of pendulum is reduced more for cone without ring and so the amplitude decreases more for cone without ring.

    Another possibility might be:
    1 mark for explaining what damping is.

    I think it's unlikely there'll be a mark for saying T is unchanged because by the word of the book 'light damping does not change the time period' (owtte), so may be considered irrelevant to what the question asks. Having said that, I often find mark schemes give marks for things that don't seem directly relevant to answering the question.
    Offline

    3
    ReputationRep:
    (Original post by NEWT0N)
    Agreed. They said "average emf" so you just subtract the end points and divide by time
    Lol I worked out V rms :/
    Offline

    2
    ReputationRep:
    (Original post by Protoxylic)
    Yes, so a quarter of a sin wave. The EMF varies over that time from 0 to a maximum emf at 0.5s. Prove for yourself that if the flux linkage is ACos(wt) for constant A=BAN. Then EMF=-dN(phi)/dt which is AwSin(wt) the emf is not constant. Hence the need to find the mean emf
    Isn't average when you just subtract the two endpoints? Like average velocity

    Also I'm fairly sure that there was a past MC question which involved a similar calculation and I remember getting it right by just doing what I did for this one
    Offline

    2
    ReputationRep:
    (Original post by CD223)
    Guys. Bad news I'm afraid

    Name:  ImageUploadedByStudent Room1434032804.520422.jpg
Views: 1179
Size:  122.5 KB


    Posted from TSR Mobile
    i did the question as if it was 150k as stated earlier meaning i did a curve but it was half way down lol is that right?
    Offline

    14
    ReputationRep:
    (Original post by NEWT0N)
    Isn't average when you just subtract the two endpoints? Like average velocity

    Also I'm fairly sure that there was a past MC question which involved a similar calculation and I remember getting it right by just doing what I did for this one
    Average velocity for CONSTANT ACCELERATION. The emf is not freaking constant it varies with sin. How many times do I actually have to explain this to people lol.

    Look at it another way, lets say a you move from 0 to 20 ms^-1 in a time of 20 seconds. Is half way in distance at t=10s? Of course it isn't
    Offline

    1
    ReputationRep:
    (Original post by aprocrastinator)
    uhhh ok, I'm sure I put my intended answer at the time lol but I ticked the middle box I believe...?

    Actually I'm pretty sure I was asking for the one with the most damping. Cos I remember I ticked the box which was the longest sentence so it would've been 'cone without the ring' ... dear god this is confusing

    I think I ticked the middle box as well.
 
 
 
Poll
Should MenACWY vaccination be compulsory at uni?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.