It's me again...can you help me with a couple of problems?

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Hexaneandheels
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Here are two exam questions I've become stuck on and it's driving me crazy. I've written in the mark scheme. Anyone able to break them down? I used to be good at physics aghast has happened!!
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langlitz
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(Original post by Hexaneandheels)
Here are two exam questions I've become stuck on and it's driving me crazy. I've written in the mark scheme. Anyone able to break them down? I used to be good at physics aghast has happened!!
For problem one:
By inspection, after 20 swings on the first pendulum, the second pendulum will be on it's 19th swing. Thus after 38 seconds they will be in phase again

For problem two:

The force on object with some mass, say m, at a distance y from M is equal to the force from the 4M

i.e. \frac{GMm}{y^2}=\frac{G4Mm}{(d-y)^2}
\frac{1}{y^2}=\frac{4}{(d-y)^2}
d^2-2dy-3y^2=0
Solve for d using the quadratic formula and you should find:
d=3y
\frac{y}{d}=\frac{1}{3}
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Hexaneandheels
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Thanks so much! The second problem makes total sense and I was actually attempting it that way but stopped myself. But the first one just doesn't make sense to my brain why can't I wrap my head around it??!
(Original post by langlitz)
For problem one:
By inspection, after 20 swings on the first pendulum, the second pendulum will be on it's 19th swing. Thus after 38 seconds they will be in phase again

For problem two:

The force on object with some mass, say m, at a distance y from M is equal to the force from the 4M

i.e. \frac{GMm}{y^2}=\frac{G4Mm}{(d-y)^2}
\frac{1}{y^2}=\frac{4}{(d-y)^2}
d^2-2dy-3y^2=0
Solve for d using the quadratic formula and you should find:
d=3y
\frac{y}{d}=\frac{1}{3}
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Callicious
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Heyo!

Ok, thing of it like this for the first one (I had some homework on it and got it wrong, but in the end it made sense, so I might not be entirely factual but yah);

The first pendulum will swing for 1.9 seconds, and will reach its start 0.1 seconds ahead of the second. After two swings, it'll be 0.2 seconds ahead, 3 swings 0.3, 4 swings? 0.4.

The second one is 0.1 behind, and it'll be 0.1 behind for the first, 0.2 for the second, 0.3 for the third, 0.4 for the fourth. After 5 swings, we can see that the 1.9 second pendulum is approximately 0.5 seconds out of phase with the second one, in terms of being ahead, and relatively the 2 second pendulum is 0.5 seconds behind. So, they're not in phase. So, yah, we can see a pattern arising here. We need to find a time where they are in phase.

We've taken 0.1 seconds as the phase difference, after all 2-1.9 is 0.1. Each swing, Pendulum 1 is 0.1 seconds ahead of Pendulum 2. We need to figure out how many of these out-of-phase revolutions it'll take. So, if you've followed me so far, you can see that after 20 revolutions of pendulum 1, we'd be 20*0.1 seconds out, 2 seconds, which is in fact the time period of the second one; we're in phase.

Now, you can use pendulum two for this too. Pendulum 2 is 0.1 seconds behind, so, for each oscillation it'll be 0.1 seconds behind the first one. After 19 revolutions, we can see that pendulum 2 is in fact 1.9 seconds behind, which is an entire revolution of pendulum 1, which means it's in phase.

I'm so frickin' sorry for my terrible formatting, etc, but I hope I helped. Tell me if you want an analogy to help you with this, because I know I make no sense.
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langlitz
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(Original post by Callicious)
Heyo!

Ok, thing of it like this for the first one (I had some homework on it and got it wrong, but in the end it made sense, so I might not be entirely factual but yah);

The first pendulum will swing for 1.9 seconds, and will reach its start 0.1 seconds ahead of the second. After two swings, it'll be 0.2 seconds ahead, 3 swings 0.3, 4 swings? 0.4.

The second one is 0.1 behind, and it'll be 0.1 behind for the first, 0.2 for the second, 0.3 for the third, 0.4 for the fourth. After 5 swings, we can see that the 1.9 second pendulum is approximately 0.5 seconds out of phase with the second one, in terms of being ahead, and relatively the 2 second pendulum is 0.5 seconds behind. So, they're not in phase. So, yah, we can see a pattern arising here. We need to find a time where they are in phase.

We've taken 0.1 seconds as the phase difference, after all 2-1.9 is 0.1. Each swing, Pendulum 1 is 0.1 seconds ahead of Pendulum 2. We need to figure out how many of these out-of-phase revolutions it'll take. So, if you've followed me so far, you can see that after 20 revolutions of pendulum 1, we'd be 20*0.1 seconds out, 2 seconds, which is in fact the time period of the second one; we're in phase.

Now, you can use pendulum two for this too. Pendulum 2 is 0.1 seconds behind, so, for each oscillation it'll be 0.1 seconds behind the first one. After 19 revolutions, we can see that pendulum 2 is in fact 1.9 seconds behind, which is an entire revolution of pendulum 1, which means it's in phase.

I'm so frickin' sorry for my terrible formatting, etc, but I hope I helped. Tell me if you want an analogy to help you with this, because I know I make no sense.
Lol.
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Oliog
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(Original post by Hexaneandheels)
Here are two exam questions I've become stuck on and it's driving me crazy. I've written in the mark scheme. Anyone able to break them down? I used to be good at physics aghast has happened!!
very tricky question this is my good man, but the answer is quite simple. if you think about the prospect that jet fuel cant melt steel beams, and the fact the 2,000 jews took the day of work that day. the only possible answer is that Bush did infact do 911. hope this is helpful fella.
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Vikingninja
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(Original post by Callicious)
Heyo!

Ok, thing of it like this for the first one (I had some homework on it and got it wrong, but in the end it made sense, so I might not be entirely factual but yah);

The first pendulum will swing for 1.9 seconds, and will reach its start 0.1 seconds ahead of the second. After two swings, it'll be 0.2 seconds ahead, 3 swings 0.3, 4 swings? 0.4.

The second one is 0.1 behind, and it'll be 0.1 behind for the first, 0.2 for the second, 0.3 for the third, 0.4 for the fourth. After 5 swings, we can see that the 1.9 second pendulum is approximately 0.5 seconds out of phase with the second one, in terms of being ahead, and relatively the 2 second pendulum is 0.5 seconds behind. So, they're not in phase. So, yah, we can see a pattern arising here. We need to find a time where they are in phase.

We've taken 0.1 seconds as the phase difference, after all 2-1.9 is 0.1. Each swing, Pendulum 1 is 0.1 seconds ahead of Pendulum 2. We need to figure out how many of these out-of-phase revolutions it'll take. So, if you've followed me so far, you can see that after 20 revolutions of pendulum 1, we'd be 20*0.1 seconds out, 2 seconds, which is in fact the time period of the second one; we're in phase.

Now, you can use pendulum two for this too. Pendulum 2 is 0.1 seconds behind, so, for each oscillation it'll be 0.1 seconds behind the first one. After 19 revolutions, we can see that pendulum 2 is in fact 1.9 seconds behind, which is an entire revolution of pendulum 1, which means it's in phase.

I'm so frickin' sorry for my terrible formatting, etc, but I hope I helped. Tell me if you want an analogy to help you with this, because I know I make no sense.
Not like as if you aren't nearly 2 years late.
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Callicious
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(Original post by Vikingninja)
Not like as if you aren't nearly 2 years late.
2 years phase difference omw
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