Who has Catholic secondary schools Trail HSC 2003 (1 Viewer)

[ressul]

Who has Catholic secondary schools Trail HSC 2003 paper? 3 unit or 4 unit.
Or does anyone know where can find it?

 

[abdo]

i have it

 

[CrashOveride]

Where do you get it from? I dont see any recent stuff for 3unit on BOS resources.

BTW: Abdo, are you the same person as that "abdooo" guy that was on here ages ago? Was wondering where he/you went to

 

[aNg3L_iReNe]

mee tooo hahaha ............... ~~

 

[coroneos]

wat the.. it is on bos!!

 

[CrashOveride]

Yes it;s on BOS.

I question from the paper itself:

A particle moving in a striahgt line is performing SHM. At time t seconds it has displacement x metres from a fixed point O on the line, velocity v m/s and acceleration x'' m/s^2 given by x'' = -4(x-2). The particle is at rest at the point O.

Show v² = -4x² + 16x

Perhaps its a bit late...but i cant see how to derive a condition for this question? It's clear that O is actually an extremity point. The suggested condition is when x=0, v=0

 

[CrashOveride]

Ok i noticed this a few times now so either im missing some completely obvious or :/

 

[Jase]

condition? what exactly do you mean by that?
the initial values of t ,x ,v or the equation?

 

[coca cola]

Yes it;s on BOS.

I question from the paper itself:

A particle moving in a striahgt line is performing SHM. At time t seconds it has displacement x metres from a fixed point O on the line, velocity v m/s and acceleration x'' m/s^2 given by x'' = -4(x-2). The particle is at rest at the point O.

Show v² = -4x² + 16x

Perhaps its a bit late...but i cant see how to derive a condition for this question? It's clear that O is actually an extremity point. The suggested condition is when x=0, v=0

what do you mean?

v² = -4x² + 16x + c, if you integrate the acceleration.

now O is the fixed point where acceleration = 0.

therefore by using acceleration equation given, 0 = -4(x-2), x must be 2.

now substitute x into velocity equation which was derived earlier, c = 0.

hence v² = -4x² + 16x.

do you do physics? it seems you've overlooked newton's first and second law of motion.

 

[CrashOveride]

O is the fixed point where velocity, not accleration, is zero (at rest means 0 velocity)

We know x=2 is centre because thats where accleration is zero, and its given its moving in SHM.

We know velocity is zero at the extremity points, so O must be one of the extremity points.

I don't see how physics can come in here, these kinds of questions are usually quite straight forard. I don't get there condition when x=0, v=0. Where did they derive this from

 

[coca cola]

O is the fixed point where velocity, not accleration, is zero (at rest means 0 velocity)

We know x=2 is centre because thats where accleration is zero, and its given its moving in SHM.

We know velocity is zero at the extremity points, so O must be one of the extremity points.

I don't see how physics can come in here, these kinds of questions are usually quite straight forard. I don't get there condition when x=0, v=0. Where did they derive this from

newton's first law of inertia equilibrium: a body acted on by no net force moves with constant velocity.

this implies that if sumF = ma = 0, velocity is constant. we know that the velocity of the particle at O is 0. this implies that the velocity is constant or was constant at O. hence we conclude by assuming the particle is matter therefore aceleration must equals to zero.

do you agree?

now the initial conditions are v=0, a=0, and x=2 as i've derived earlier. is this enough conditions for you to solve the problem?

 

[CrashOveride]

You say at x=2 accel is zero (correct, no accel at center)
but u say velocity is also zero. How can zero velocity be achieved at the center of oscilatiion for an object moving in SHM. The centre of osc. is where max velocity occurs.

So im still wondering about the suggested solution conditions (x=0, v=0) which would enable me to solve the problem.

 

[mr EaZy]

i have it

U legend!!

 

[CM_Tutor]

now the initial conditions are v=0, a=0, and x=2 as i've derived earlier. is this enough conditions for you to solve the problem?

Coca Cola, an object with v = 0 and a = 0 is stationary and will remain so. It isn't moving in SHM because it isn't moving at all.

In the problem given, at the centre (x = 2), a = 0 but v ≠ 0.

 

[CrashOveride]

CM_Tutor: I dont think it;s possible to work out this question? We need more info ?

 

[CM_Tutor]

CrashOveride - I don't see the problem. We have x'' = -4(x - 2), and the particle is at rest at O.

x'' = d/dx(v² / 2) = -4x + 8
v² / 2 = ∫ -4x + 8 dx
v² / 2 = -(4x² / 2) + 8x + C

Now, at x = 0, v = 0: 0 = 0 + 0 + C
So, C = 0

So, v² = -4x² + 16x, as required.

Am I missing something?

 

[CrashOveride]

CrashOveride - I don't see the problem. We have x'' = -4(x - 2), and the particle is at rest at O.

x'' = d/dx(v2 / 2) = -4x + 8
v2 / 2 = ∫ -4x + 8 dx
v2 / 2 = -(4x2 / 2) + 8x + C

Now, at x = 0, v = 0: 0 = 0 + 0 + C
So, C = 0

So, v2 = -4x2 + 16x, as required.

Am I missing something?

You have said at x=0, v=0. I assume you have infered O to be the centre of motion? But we can see from the acceleration equation given above that x=2 is actually the centre of oscillation. If I utilize the v² = n²(a² - x² and input v=0 we see that the point O is actually the extremity point - and we also know as common knowledge that the velocity is zero(particle is at rest) at the extremity points.

You have taken the same approach as the solutions, and I don't believe it's correct.