# Oxford Physics: PAT test discussionWatch

1 year ago
#201
Y equation was made of omega*t also. So, if u saw both the equations in terms of t and not omega*t... It won't make a difference.

See... Let's say the X equation was a(t - sine(t)) sine(t) will take the value of 1/2 ONLY. But.... t can take MANY values: pi/6, pi - pi/6, 2pi + pi/6 are some examples. (note that The sine of all these values is the same: 1/2)
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1 year ago
#202
Y equation was made of omega*t also. So, if u saw both the equations in terms of t and not omega*t... It won't make a difference.

See... Let's say the X equation was a(t - sine(t)) sine(t) will take the value of 1/2 ONLY. But.... t can take MANY values: pi/6, pi - pi/6, 2pi + pi/6 are some examples. (note that The sine of all these values is the same: 1/2)
Yes indeed, but was everwhere where i wrote t omega t? Becouse than the solution would be the same
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1 year ago
#203
(Original post by Tomyil12345)
Yes indeed, but was everwhere where i wrote t omega t? Becouse than the solution would be the same
But the solution of root3-2cosx is x is +- 1/6 pi so when you plug this into sine you get +-1/2
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1 year ago
#204
(Original post by Tomyil12345)
Yes indeed, but was everwhere where i wrote t omega t? Becouse than the solution would be the same
Yep!
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1 year ago
#205
for the circles question, if I got one of the radii wrong but the centres of both and the other radius right and correctly drew in all four tangents but was unable to work out their lengths, how many marks do you think that would be?
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1 year ago
#206
(Original post by Tomyil12345)
But the solution of root3-2cosx is x is +- 1/6 pi so when you plug this into sine you get +-1/2
Oh yeah... Then there are 4 sets of solutions (general solutions).
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1 year ago
#207
(Original post by Quantum42)
for the circles question, if I got one of the radii wrong but the centres of both and the other radius right and correctly drew in all four tangents but was unable to work out their lengths, how many marks do you think that would be?
How much did you get the radii as?

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1 year ago
#208
Now let's address the most crazy thing - waves question! What was that?!?! How the hell would you GRAPH such a thing?

Also, can anyone post answers to the other questions involved in the wave question? (Time period, velocity, frequency related)
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1 year ago
#209
Okay here is the full and I believe correct solution to the binary system:

For the 2 star binary system, notice that the graviational force on a star is
, this is because the gravitational force between the 2 stars is the inversely proportion to the square of the distance BETWEEN THEM, which in this case is , notice that the radius of orbit of a star is so we have the centripetal force is . So after rearranging we have that . The distance squared between 2 stars in the 3 star system is 3, so the graviational force is , resolving this force towards the center and recognising that there are 2 stars we get that the centripetal force is , we finally get that . I'll leave it as an exercise for you to find the ratio.
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1 year ago
#210
How much did you get the radii as?

2.5 and 3
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1 year ago
#211
(Original post by Quantum42)
2.5 and 3
I got the radii as 5 and 3. I am pretty sure I didn't make a mistake, but everyone here seems to be getting 2.5 and 1.5. Idk.
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1 year ago
#212
Oh yeah... Then there are 4 sets of solutions (general solutions).
No 2 i think
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1 year ago
#213
I got the radii as 5 and 3. I am pretty sure I didn't make a mistake, but everyone here seems to be getting 2.5 and 1.5. Idk.
I've heard a mixture of both
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1 year ago
#214
(Original post by Electric-man7)
Okay here is the full and I believe correct solution to the binary system:

For the 2 star binary system, notice that the graviational force on a star is
, this is because the gravitational force between the 2 stars is the inversely proportion to the square of the distance BETWEEN THEM, which in this case is , notice that the radius of orbit of a star is so we have the centripetal force is . So after rearranging we have that . The distance squared between 2 stars in the 3 star system is 3, so the graviational force is , resolving this force towards the center and recognising that there are 2 stars we get that the centripetal force is , we finally get that . I'll leave it as an exercise for you to find the ratio.
Exactly the answer I wrote before. V_3 = 2*(1/3)^(1/4)*v_2
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1 year ago
#215
(Original post by Tomyil12345)
No 2 i think
2 sets of sine solutions with 1/2 and 2 sets with -1/2. Of course it can be totally 2 sets depending on how u write the general solution (U probably write both the 2pi + theta and pi - theta as a combined general solution. In that case, 2 sets)
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1 year ago
#216
Exactly the answer I wrote before. V_3 = 2*(1/3)^(1/4)*v_2
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1 year ago
#217
2 sets of sine solutions with 1/2 and 2 sets with -1/2. Of course it can be totally 2 sets depending on how u write the general solution (U probably write both the 2pi + theta and pi - theta as a combined general solution. In that case, 2 sets)
No because with -1/6 pi comes - 1/2 and 1/6pi vice versa so you get two general

And the solution for root3/2 =cos x is x
=+- 1/6pi pi- theta is for sine x = root 3/2
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1 year ago
#218
Guys predicted shortlist threshold?
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1 year ago
#219
(Original post by Tomyil12345)
No becouse with -1/6 pi comes - 1/2 and 1/6pi vice versa so you get two general
Oh... U mean to say the in the 4 general ones that I am talking about, 2 will be redundant cuz they'll be the same? Yeah, I guess.
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1 year ago
#220
Guys I if managed to find the equations of the circles and drew them well, and drew the tangents but could not find their lenghts how much will I get?
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