Simple harmomic motion

1. Mar 7, 2007

imy786

1. The problem statement, all variables and given/known data

amplitude is kept at 4.0 cm and the vibration
frequency increased to 50 Hz. T
The particle passes through the origin at time t = 0.0 s, travelling in the +y-direction.

(i) At what time does the valve first pass through the position y = 3.0 cm?
(ii) What is the speed at this point?
(iii) What is the acceleration at this point?
(iv) At what positions will the valve have a speed of 6.0ms−1?

2. Relevant equations

w= 2*pi*f
acceleration= -w^2*x
v= rw

3. The attempt at a solution

w= 2*pi*f
= 2*50*pi= 100pi

acceleration= w^2*x
= 100pi*4= 400pi

dont know if this is correct?
also dont know how to get the time and postions part (i and iv)

2. Mar 7, 2007

dimensionless

What is the equation for the motion?

What do you call the time derivative of the equation of motion?

Last edited: Mar 7, 2007
3. Mar 7, 2007

dobry_den

ad i) Do you know the equation for simple harmonic motion - i.e. the relation between position and time? Solving this part is just a matter of solving this equation for y = 3.

ad iv) The same story. Just use the equation* for velocity of simple harmonic movement and solve for v = 6 m/s to get the time and use the equation in i) to get the position.

*you might get this one by differentiating the equation in i)

don't forget the minus sign - do you know what its significance is?

Last edited: Mar 7, 2007
4. Mar 14, 2007

imy786

i)

equation of motion
v2= u2+2as

w= 2pi f
w= 100 pi

equation for SHM:x= asin wt
3= 4 sin 100 pi t
0.75= sin 100pi t
inverse 0.75= 100pi t
48.590...= 100 pi t
t= 1.527 secs

5. Mar 14, 2007

imy786

- indicates acceleartion towards the starting position

6. Mar 14, 2007

imy786

ii)
v= rw
= 3pi

7. Mar 14, 2007

imy786

iv)

a= vw
= 600pi

is questions i- iv correct???

8. Mar 14, 2007

Mentz114

i) is correct.
ii) is wrong. To get a velocity differentiate the x=asinwt wrt t.

9. Mar 14, 2007

imy786

ii)v= awcos wt

a= 3
w= 100pi
t=0
using this values would this be ok ?
iii)
iv)

10. Mar 15, 2007

imy786

ii)

coz0= 1
v= 300 pi

is this correct?

11. Mar 15, 2007

nithin

Ok for the first one , you need to form a equation and then solve it . It is given that the y - position is 0 at t=0. w=100pi and A=0.04m (always work with s.i units) . Using these info we can form the equation , y=A Sin(wt) , in this case it would be 0.03=0.04Sin((100pi)(t)) , solving for this will result in the answer!!!

12. Mar 15, 2007

nithin

Last edited: Mar 15, 2007
13. Mar 15, 2007

nithin

For part 2 , once again we need to form a equation, this time , we need to form the equation for velocity. Which would be v=-wACos(wt) (it is cos because at origin , y=0) From i) you have your t which you can sustitute into the equation and solve it!!

Last edited: Mar 15, 2007
14. Mar 15, 2007

nithin

Ok for part 3) , you need to form another equation. The equation would be a=-W^2ASin(wt). Using the ans from part 1) , you can sole the equation and get your ans!!!

15. Mar 15, 2007

nithin

just for you info imy786 , v=Aw and a= Aw^2 give you the maximum velocity and maximum acceleration.

16. Mar 15, 2007

imy786

nithin....

v=-wACos(wt

i dont think it has a minus sign.

book quotes

v= Awcos wt

17. Mar 15, 2007

imy786

also

a= Aw^2 sin (wt)
v= Awcos wt

you wrote these equations -

v=-wACos(wt
a=-W^2ASin(wt

18. Mar 15, 2007

imy786

URGENT NEED help............

19. Mar 15, 2007

nithin

imy786 it depends on the situation of the question . Sketch a simple y vs t and v vs t graph side byside ,You will then see whether it is - or +

20. Mar 15, 2007

nithin

I use cos instead of sin in my equations as my phase angle is 0 as it starts from y=0 at t=0. If u wish u can also use y= A Sin(wt + phi) then v= -AWCos(wt + phi) .Ooops my bad . Sorry .Like i said just draw a graph and you will see what to do.