AQA M2 Differential Equations Watch

maggiehodgson
Badges: 14
Rep:
?
#1
Report Thread starter 4 years ago
#1
Hi

I'm on question 5. The answer I'm getting is not the one in the book

\frac{d^{2}x}{dt^{2}}= \frac{-10x}{3}

I can see that the particle will slow down so I understand the negative. I don't know why it's x and not x-3.

I've not used I've not used the initial speed nor the 2kg . Does that just come in to part b) which I've not had a go at yet or am I missing something from part a)?

Thanks
Attached files
0
reply
ghostwalker
  • Study Helper
Badges: 16
#2
Report 4 years ago
#2
[QUOTE=maggiehodgson;51325497]Hi

I'm on question 5. The answer I'm getting is not the one in the book

\frac{d^{2}x}{dt^{2}}= \frac{-10x}{3}

2nd, corrected, attempt.

The book is wrong, IMO.

I can see that the particle will slow down so I understand the negative. I don't know why it's x and not x-3.
It's negative because the force is in the opposite direction to x increasing.

I've not used I've not used the initial speed nor the 2kg . Does that just come in to part b) which I've not had a go at yet or am I missing something from part a)?

Thanks
Well you used the 2kg in part a).

The initial speed doesn't come into it at all.

I suspect the question in the book was modified without being checked before they went to press.
reply
maggiehodgson
Badges: 14
Rep:
?
#3
Report Thread starter 4 years ago
#3
[QUOTE=ghostwalker;51326593]
(Original post by maggiehodgson)
Hi

I'm on question 5. The answer I'm getting is not the one in the book

\frac{d^{2}x}{dt^{2}}= \frac{-10x}{3}

2nd, corrected, attempt.

The book is wrong, IMO.



It's negative because the force is in the opposite direction to x increasing.



Well you used the 2kg in part a).

The initial speed doesn't come into it at all.

I suspect the question in the book was modified without being checked before they went to press.
Ok. Thanks for the help. Yes, of course I used the 2kg -what was I thinking!


For part b) the answer is
\frac{d^{2}x}{dt^{2}}= \frac{-10x}{3} \mp 0.1g
where the sign is + if
\frac{dx}{dt}<0}}
and the sign is - if
\frac{dx}{dt}>0}}

When the movement is to the right (as in my diagram) the velocity is positive and when the particle is being pulled back the velocity is negative? Does that explain the +/-?

Where does the 0.1g come from or is the book answer all wrapped up in being a modified question?
0
reply
tiny hobbit
Badges: 15
Rep:
?
#4
Report 4 years ago
#4
[QUOTE=maggiehodgson;51328677]
(Original post by ghostwalker)

Ok. Thanks for the help. Yes, of course I used the 2kg -what was I thinking!


For part b) the answer is
\frac{d^{2}x}{dt^{2}}= \frac{-10x}{3} \mp 0.1g
where the sign is + if
\frac{dx}{dt}<0}}
and the sign is - if
\frac{dx}{dt}>0}}

When the movement is to the right (as in my diagram) the velocity is positive and when the particle is being pulled back the velocity is negative? Does that explain the +/-?

Where does the 0.1g come from or is the book answer all wrapped up in being a modified question?
In part b), you now have to include a frictional force in your equation. If the particle is moving in the positive direction, the friction is acting to the left like the tension. Friction = mu x reaction, reaction = weight and then the 2 will cancel out. Give it a go!
0
reply
maggiehodgson
Badges: 14
Rep:
?
#5
Report Thread starter 4 years ago
#5
[QUOTE=tiny hobbit;51329207]
(Original post by maggiehodgson)

In part b), you now have to include a frictional force in your equation. If the particle is moving in the positive direction, the friction is acting to the left like the tension. Friction = mu x reaction, reaction = weight and then the 2 will cancel out. Give it a go!
R = mg = 2g
F = muR =0.1x2g

How does the 2 cancel out? I must be in a really thick mode today.
0
reply
ghostwalker
  • Study Helper
Badges: 16
#6
Report 4 years ago
#6
(Original post by maggiehodgson)

R = mg = 2g
F = muR =0.1x2g

How does the 2 cancel out? I must be in a really thick mode today.
F= 0.2g

Then resultant force is -20x/3 +/- 0.2g and this equals ma = 2a and 2's cancel throughout.
reply
maggiehodgson
Badges: 14
Rep:
?
#7
Report Thread starter 4 years ago
#7
(Original post by ghostwalker)
F= 0.2g

Then resultant force is -20x/3 +/- 0.2g and this equals ma = 2a and 2's cancel throughout.

Yep. Got it.

Just to be clear it is -20(x-3)/3 isn't it not -20x/3?
0
reply
ghostwalker
  • Study Helper
Badges: 16
#8
Report 4 years ago
#8
(Original post by maggiehodgson)
Yep. Got it.

Just to be clear it is -20(x-3)/3 isn't it not -20x/3?
Yes.
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

University open days

  • Cranfield University
    Cranfield Forensic MSc Programme Open Day Postgraduate
    Thu, 25 Apr '19
  • University of the Arts London
    Open day: MA Footwear and MA Fashion Artefact Postgraduate
    Thu, 25 Apr '19
  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 27 Apr '19

Have you registered to vote?

Yes! (443)
37.86%
No - but I will (88)
7.52%
No - I don't want to (80)
6.84%
No - I can't vote (<18, not in UK, etc) (559)
47.78%

Watched Threads

View All
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