AQA Physics A - PHYA4 (11/06/12) - Exam thread

Guy's what is meant by a mechanical system?

E.g. A mechanical system is oscillating at resonance with a constant amplitude. Which one of the following statements is not correct?

A. The applied force prevents the amplitude from becoming too large.
B. The freq. of the applied force is the same as the natural freq. of the oscillation of the system.
C. The total energy of the system is constant.
D. The amplitude of oscillations depends on the amount of damping.

For this question, I know B and D are correct which leaves A and C. I don't get what A is saying, but for C surely there will be energy losses due to air resistance so the total energy of the system is not constant?

Whats the right answer? i think its B that isnt correct, as if it was at resonance the amplitude would continue to increase very high

Reply 181

schizopear

Original post by Next Level

Guy's what is meant by a mechanical system?

E.g. A mechanical system is oscillating at resonance with a constant amplitude. Which one of the following statements is not correct?

A. The applied force prevents the amplitude from becoming too large.
B. The freq. of the applied force is the same as the natural freq. of the oscillation of the system.
C. The total energy of the system is constant.
D. The amplitude of oscillations depends on the amount of damping.

For this question, I know B and D are correct which leaves A and C. I don't get what A is saying, but for C surely there will be energy losses due to air resistance so the total energy of the system is not constant?

i would have gone with A, because if the system is resonating then the amplitude is maximum - so the applied force is causing the amplitude to be large not preventing it?

Reply 182

m1a1tank

1

It is A. Because it is at resonance the frequency of the applied force has to be equal to the natural frequency therefore B has to be correct. The question is asking for what is INCORRECT.

Like the previous person said. Its oscillating at resonance so if anything the applied force is what is making the amplitude become too large right. A is saying the opposite to that so A is the statement that is INCORRECT.

Reply 183

Next Level

1

Original post by schizopear

i would have gone with A, because if the system is resonating then the amplitude is maximum - so the applied force is causing the amplitude to be large not preventing it?

Yeah A is right, makes sense too sorry was just finding it hard to understand the statement of A :tongue:

Cheers guys!

Reply 184

sconter

11

http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4-1-W-QP-JAN11.PDF can u help me with 11 please

Reply 185

BEB21

Can anyone help me with http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN11.PDF question 16 and 21 please?

Reply 186

FrightBright

Original post by BEB21

Can anyone help me with http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN11.PDF question 16 and 21 please?

Link dont work but I checked out the paper anyway.

For 16 you just use F=GMm/r^2 and E=Qq/4(pi)(epsilon naught)

You get the two forces and divide them by each other, that should give you the ratio.

For 21

Well it turns anti clockwise until vertical, because thetha = 0

So max Flux linkage is BAN.

BANcos50 will give you previous flux linkage.

Then subtract them... Its obviously increasing because its cutting more field lines.

Reply 187

FrightBright

Original post by sconter

http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4-1-W-QP-JAN11.PDF can u help me with 11 please

This has been answered to death. Flick through the thread.

To cut it short make mw^2r = mg.

Reply 188

justravi

1

Original post by sconter

http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4-1-W-QP-JAN11.PDF can u help me with 11 please

Haven't looked at the answer but it should be right.

Assuming it's B, 1.4 hours:

Equate the equations for "magnitude of gravitational field strength in radial field" and "centripetal acceleration" (the one with angular velocity squared * r) and make T the subject which is 1/f.

thanks both of u.

Original post by FrightBright

Link dont work but I checked out the paper anyway.

For 16 you just use F=GMm/r^2 and E=Qq/4(pi)(epsilon naught)

You get the two forces and divide them by each other, that should give you the ratio.

For 21

Well it turns anti clockwise until vertical, because thetha = 0

So max Flux linkage is BAN.

BANcos50 will give you previous flux linkage.

Then subtract them... Its obviously increasing because its cutting more field lines.

Thank you. I get 21 now, but 16...I try and do it like you say but I still get around 10^30 when it should be 10^36...I think I'm missing something really obvious!

Reply 191

radiator0505

How do you explain why a charged particle moves in a circle when its in a magnetic field?
I know it has something to do with force being perpendicular to velocity but I don't know how to word it.

Reply 192

don'tTRIP.

2

Original post by radiator0505

How do you explain why a charged particle moves in a circle when its in a magnetic field?
I know it has something to do with force being perpendicular to velocity but I don't know how to word it.

Using Fleming's left-hand rule, force is perpendicular to velocity. The force changes the direction of the velocity, but not its magnitude; hence it moves in a circle. Probably a better answer out there tbh :holmes:

Reply 193

number23

15

Original post by radiator0505

How do you explain why a charged particle moves in a circle when its in a magnetic field?
I know it has something to do with force being perpendicular to velocity but I don't know how to word it.

the magnetic force acts perpendicular to the line of motion, acting as a centripetal force, causing circular motion

Reply 194

number23

15

Original post by don'tTRIP.

Using Fleming's left-hand rule, force is perpendicular to velocity. The force changes the direction of the velocity, but not its magnitude; hence it moves in a circle. Probably a better answer out there tbh :holmes:

isnt flemings LHR for force, field and current... it doesnt involve velocity does it? :confused:

Reply 195

don'tTRIP.

2

Original post by number23

isnt flemings LHR for force, field and current... it doesnt involve velocity does it? :confused:

I'm probably confusing current with velocity :facepalm2:

Let's hope that doesn't happen on Monday; anyone give a definitive answer?

Reply 196

radiator0505

Original post by don'tTRIP.

I'm probably confusing current with velocity :facepalm2:

Let's hope that doesn't happen on Monday; anyone give a definitive answer?

Flemmings rule works for F=BIL and F=BQv, because:
F=BIL, I=Q/t
F=B(Q/t)L
F=BQ(L/t)
L/t=v
F=BQV

Reply 197

FrightBright

Original post by don'tTRIP.

I'm probably confusing current with velocity :facepalm2:

Let's hope that doesn't happen on Monday; anyone give a definitive answer?

The direction of current and direction of velocity is the same thing.

Reply 198

don'tTRIP.

2

If we're given the charge-to-mass ratio of an ion and nothing else, how can we work out the electric field strength? E/m = 2.4 x 10^7; electric field strength = 4.09 x 10^-7