Oxford Physics: PAT test discussion Watch

cooltejaskd
Badges: 8
Rep:
?
#201
Report 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)
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
Tomyil12345
Badges: 9
Rep:
?
#202
Report 1 year ago
#202
(Original post by cooltejaskd)
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
0
quote
reply
Tomyil12345
Badges: 9
Rep:
?
#203
Report 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
0
quote
reply
cooltejaskd
Badges: 8
Rep:
?
#204
Report 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!
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
Quantum42
Badges: 13
Rep:
?
#205
Report 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?
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
cooltejaskd
Badges: 8
Rep:
?
#206
Report 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).
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
cooltejaskd
Badges: 8
Rep:
?
#207
Report 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?

To answer your question, I think very less. Probably 2/9.
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
cooltejaskd
Badges: 8
Rep:
?
#208
Report 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)
Posted on the TSR App. Download from Apple or Google Play
1
quote
reply
Electric-man7
Badges: 8
Rep:
?
#209
Report 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
 \frac{Gm^2}{4R^2}, 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 2R, notice that the radius of orbit of a star is R so we have the centripetal force is \frac{mv_2^2}{R}. So after rearranging we have that {v_2^2}=\frac{Gm}{4R}. The distance squared between 2 stars in the 3 star system is 3, so the graviational force is \frac{Gm^2}{3R^2}, resolving this force towards the center and recognising that there are 2 stars we get that the centripetal force is \frac{\sqrt{3}Gm^2}{3R^2}, we finally get that v_3^2=\frac{\sqrt{3}Gm}{3R}. I'll leave it as an exercise for you to find the ratio.
0
quote
reply
Quantum42
Badges: 13
Rep:
?
#210
Report 1 year ago
#210
(Original post by cooltejaskd)
How much did you get the radii as?

To answer your question, I think very less. Probably 2/9.
2.5 and 3
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
cooltejaskd
Badges: 8
Rep:
?
#211
Report 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.
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
Tomyil12345
Badges: 9
Rep:
?
#212
Report 1 year ago
#212
(Original post by cooltejaskd)
Oh yeah... Then there are 4 sets of solutions (general solutions).
No 2 i think
0
quote
reply
Quantum42
Badges: 13
Rep:
?
#213
Report 1 year ago
#213
(Original post by cooltejaskd)
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
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
cooltejaskd
Badges: 8
Rep:
?
#214
Report 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
 \frac{Gm^2}{4R^2}, 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 2R, notice that the radius of orbit of a star is R so we have the centripetal force is \frac{mv_2^2}{R}. So after rearranging we have that {v_2^2}=\frac{Gm}{4R}. The distance squared between 2 stars in the 3 star system is 3, so the graviational force is \frac{Gm^2}{3R^2}, resolving this force towards the center and recognising that there are 2 stars we get that the centripetal force is \frac{\sqrt{3}Gm^2}{3R^2}, we finally get that v_3^2=\frac{\sqrt{3}Gm}{3R}. 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
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
cooltejaskd
Badges: 8
Rep:
?
#215
Report 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)
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
Electric-man7
Badges: 8
Rep:
?
#216
Report 1 year ago
#216
(Original post by cooltejaskd)
Exactly the answer I wrote before. V_3 = 2*(1/3)^(1/4)*v_2
Oh my bad I misinterpeted your writing.
0
quote
reply
Tomyil12345
Badges: 9
Rep:
?
#217
Report 1 year ago
#217
(Original post by cooltejaskd)
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
0
quote
reply
Electric-man7
Badges: 8
Rep:
?
#218
Report 1 year ago
#218
Guys predicted shortlist threshold?
0
quote
reply
cooltejaskd
Badges: 8
Rep:
?
#219
Report 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.
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
Electric-man7
Badges: 8
Rep:
?
#220
Report 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?
0
quote
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Did you get less than your required grades and still get into university?

Yes (78)
30%
No - I got the required grades (149)
57.31%
No - I missed the required grades and didn't get in (33)
12.69%

Watched Threads

View All