Last probl. v can be found using 1st probl. v^2 = (w^2)(a^2) - (w^2)(x^2) where a=0.4 max displacement when v = 0. Then max velocity is when x = 0 i.e. v^2 = (w^2)(a^2), v = w*a = 5 * 0.4 = 2 (speed = positive sqr)

@BBK583

2 ай бұрын

just wrote this out and realised surely you cannot go from (v² = w² * a²) to (v = w * a)?? i subbed in values into (v² = w² * a²) and got: v² = 5² * 0.4² v² = 25 * 1.6 v² = 40 v is not 2 then?

Do you have any more fp1 videos for further? I am not a student at your school, but your maths videos are very very helpful

@BicenMaths

4 жыл бұрын

Unfortunately we are not teaching FP1 to our current Year 12s, and so there aren’t videos available from Year 1 content - I also only started the videos midway through the year so some of the Year 13 content is missing. There’ll be some videos on vectors coming up very soon, and before that videos on transforming differential equations, both Year 13 FP1 topics. Watch this space! Pleased the videos are helpful 👍🏼

at 34:47 can you also work out the maximum speed by setting dv/dt = 0 to get the turning point and working out the value for t and then subbing back into equation for v?

@BicenMaths

Ай бұрын

Yes you can - max speed is when acceleration is 0, which is dv/dt=0

27:25 For these types of questions, especially in mechanics, where the question doesn’t explicitly state to give me answer to a certain number of decimal places or significant figures is it okay to give my answer in its unrounded form? Would I still get all the marks? In this example, x is actually root(5) which is what I gave as my answer.

@BicenMaths

Ай бұрын

Yes, I think that should be fine - to be safe I'd write x=root5=2.24 (3sf) so that both options are there for them to see.

at 7:00 the fastest acceleration would be when the particle is going through O right? But then displacement x = 0 then so how does that make sense saying it doesn't accelerate at all at that point?

@BicenMaths

2 ай бұрын

No, the fastest acceleration is when it is furthest away from O, when the velocity is 0 - this is because the force pulling it back to the centre is largest. When it is travelling through O, the displacement is indeed 0, which means that the acceleration is 0 - which means that it is at its maximum velocity, as after is passes O, it will start to slow down again!

can I ask why when finding the max speed you put whatever value t is when x=0 into the general solution rather than differentiating the particular solution and putting said value of t there 35:18

@BicenMaths

Ай бұрын

Because I know at x=0 the speed is a maximum, as as soon as it leaves that position, it will start slowing down again!

For part (b) of the particle question, can’t we use the fact that when t=0, x=0 so 0= Pcos(0)+0 then 0=P(1) => P=0 Also maximum "x" is “a” and it happens when ωt = π/2 so: a = Q(1) so Q=a and now we have that x= asin(ωt+α)

@BicenMaths

5 ай бұрын

Yes that's also a great approach!

At 28:33 how did you know it turns into Rsin(2t+alpha) and not Rcos(2t+alpha) or Rsin(2t-alpha)? Are we allowed to use any of them? I was just curious as to why you picked the sin one because the textbook also does. I also tried it with Rcos(2t+alpha) and for me it was slightly off. I know we only needed the amplitude but I was trying to remember how to do these again.

@BicenMaths

Ай бұрын

You can use any of them - but if it has a '+ sign' I would use either Rsin(2t+alpha) or Rcos(2t-alpha) as these are the ones that have a + rather than a - sign. Either would work!

Sir, to correct you at 5:10 the "ω" in the simple harmonic motion equation refers to its "angular frequency" very similar to "angular velocity" but not quite. Angular velocity has direction and magnitude (vector) is defined as dθ/dt. Whereas angular frequency has only magnitude (scalar), note that there is no change in angle (θ), it purely measures how "often" something repeats (or any kind of periodic phenomena) i.e frequency. Your explanation about the point moving in a circle is the correct explanation of angular velocity because there's a change in angle dθ/dt, but this not correct for a spring system as the is no change in angle. It's very similar to a spring system and can be compared to each other but they are not identical. They are both calculated by 2π/T where T is the time period or 2πf where f is the frequency.

@BicenMaths

Ай бұрын

Great comment! I go into a lot more detail of omega in my FM2 playlist where this concept is explored more deeply, but you've done a great summary here!

what if it was a, let's say second order non-homogenous, differential equation; is the term in front of the x (multiplied by -1 as -omega^2) the number of oscillations per 2pi seconds?

@BicenMaths

25 күн бұрын

Yes - but they never go into that much detail in CP as it is covered in more detail in FM2

@k4ndy_067

25 күн бұрын

@@BicenMaths Alright, thanks.

28:36 how would you know its sin? or could you take either?

@BicenMaths

Жыл бұрын

You could do either!

5:38 😂😂😂

35:29 could of differentiated it to get v=-2sin5t, realise 1 ≥ |sin5t|

@BicenMaths

Ай бұрын

Yeah, excellent, love this