Shm (1 Viewer)

[CrashOveride]

acceleration of a body moving along the x-axis is given by:

d/dx(v^2/2) = -x for x<= 1
d/dx(v^2/2) = 2-x for x > 1

when t=0, x=0 and v=1/3.......determine the extremem points of subsequent motion and hence prove that the motion is SHM

Ok so what i said was: given the conditions its accelerating by way of the 1st case, it cant go anymore than 1/3 because we cant get -ve sqrt and thus will turn back....and by doing so it never exploits the 2nd case. So extreme points are x= +- 1/3.

Now part c) says if that velocity at t=0,x=0 was v=2, find the extreme points of the motion. Explain whether the motion now is SHM or not. I got a bit stuffed up cos i didnt get the extremities in the back of the book and it ends up like crossing into the other acceleration set so that mucks around with it ???

 

[CrashOveride]

Also:

P, Q, R are three points on the x-axis such that PQ = QR = 2m. A particle performs SHM along x-axis and has velocity of 12, 10 and 6 at P , Q, R respectively. Taking the origin at the centre of oscillation, find the constant n in the eqn x'' = -n²x

 

[CM_Tutor]

Case: At t = 0, v = 1 / 3. Motion governed by d/dx(v² / 2) = - x at all times. SHM on -1 / 3 &le; x &le; 1 / 3

Case: At t = 0, v = 2. Motion governed by both cases. Motion is periodic. End points are at x = 4 and x = -2. Motion is not SHM, as not symmetric about x = 1.

 

[CM_Tutor]

Also:

P, Q, R are three points on the x-axis such that PQ = QR = 2m. A particle performs SHM along x-axis and has velocity of 12, 10 and 6 at P , Q, R respectively. Taking the origin at the centre of oscillation, find the constant n in the eqn x'' = -n²x

n = &radic;10 / 2

 

[CrashOveride]

You do that off the top of your head?

PS. The previous question back of book says extreme points are 2 +- 2sqrt(2)

 

[CM_Tutor]

You do that off the top of your head?

No. I showed that P was 17 / 5 units from the origin, that the amplitude of the motion was sqrt(69.16), and hence that n² = 5 / 2

PS. The previous question back of book says extreme points are 2 +- 2sqrt(2)

I don't see where I've made a mistake, but I could be wrong. What did you get for the end points?

 

[CrashOveride]

If you dont mind could you post up some working? Because for the P Q R question i wasn;t even sure where to start, i tried doing some things finding expressions for the points as functions of time t and velocities etc. but i couldnt go any further.

 

[CM_Tutor]

Since the velocity is higher at P than at R, it follows P is nearer the centre of the motion. So, put P at x = d. It follows that Q is at x = d + 2, and R is at x = d + 4, and you know the velocities at each of these points. We also know that if x'' = -n²x, then v² = n²(A² - x²), where A is the amplitude.

So, you have three simultaneuos equations in three unknowns A, d and n, to solve for n:

144 = n²(A² - d²)
100 = n²[A² - (d + 2)²]
36 = n²[A² - (d + 4)²]

 

[CrashOveride]

Crikey, didn't even occur to me.

Ah well makes sense now.

For that other question, i got the extremities you got and i said because we have x'' = 0 at x=2, then its not SHM becasue x''=0 only at centre of oscillation.

 

[CM_Tutor]

For that other question, i got the extremities you got and i said because we have x'' = 0 at x=2, then its not SHM becasue x''=0 only at centre of oscillation.

Then maybe the book is wrong? You actually have x'' = 0 at x = 0 as well, another reason it can't be SHM.

 

[CrashOveride]

Yeah that too.