# Oscillation - acceleration

1. Apr 21, 2010

### MissPenguins

1. The problem statement, all variables and given/known data
A 0.7 kg block attached to a spring of force
constant 13.4 N/m oscillates with an ampli-
tude of 20 cm.
Find the maximum speed of the block.

(part 2 of 4) 10.0 points
Find the speed of the block when it is 10 cm
from the equilibrium position.
(part 3 of 4) 10.0 points
Find its acceleration at 10 cm from the equi-
librium position.
014 (part 4 of 4) 10.0 points
Find the time it takes the block to move from
x = 0 to x = 10 cm.

I already answered part 1, and 2. I got .875 in the first part, .7578 for the second part. But I don't know how to do part 3 and 4.

2. Relevant equations
For part 3, I tried using v = $$\pm$$$$\sqrt{}$$k/m(A2-x2)

3. The attempt at a solution
I tried to find the derivative of the above equation for acceleration. I don't think I did the derivative right. Can someone please help?

2. Apr 22, 2010

### ehild

You certainly learned Hook's law about how much force does a stretched spring exert on a body attached to its end. And you know that force = mass times acceleration. So how much is the force of the spring when it is stretched by 10 cm from its equilibrium position?

As for the time it takes the block to move from x=0 to x=10: This motion has the time dependence x = A sin (wt). You know w from the spring constant and mass. The amplitude is given: 20 cm. At t = 0, x=0. Find the argument of the sine when x/A =10/20. Divide by w.

ehild

3. Apr 22, 2010

### MissPenguins

The Force isn't given for in order to use F = ma. So I used a = -w2Acos(wt+theta). w2=k/m. I did all those and I got -70.1599. Is it right? Does it make sense?

For the second part, I used x(t) = A cos(wt), so I calculated (0.1 m)/(0.2 m) = 0.5 m. cos inverse, then I found the t to be 1.047198. Is it correct?

4. Apr 23, 2010

### ehild

You do know the force of a spring: it is F = -k(x-x0). x-x0 = 0.1 m is the deviation from the equilibrium position, k is the spring constant, k=13.4 N/m. How much is the force?

You can use also the formula a = -w2Acos(wt+theta), but I do not see how did you got 70.16. In what unit is it?

No it is not correct. You have to get the elapsed time from the equilibrium position, that is x=0, to x=0.1 m. At what time is x=0 ?

ehild

5. Apr 23, 2010

### MissPenguins

Alright, for the first part, I did F = -k(x-x0) = (- 13.4 N/m)(-0.1 m) =1.34 N.
F = ma => F/m = a, 1.34 N/ 0.7 kg = 1.914 m/s2. Is that correct?

I still don't get the other part. Thanks. :)

6. Apr 23, 2010

### ehild

You assumed that the position of block is x=A cos(wt), but it can be taken x = A sin (wt), as well. It is the same motion, only the starting time is different. If x = 0.2 sin(wt) the block is at equilibrium position at t=0. What is wt when it is at x=0.1 next? See picture.

ehild

Last edited: Jun 29, 2010
7. Apr 24, 2010

### MissPenguins

wt is at 0.5 when x = 0.1.

8. Apr 24, 2010

### Bearbull24.5

I have this same problem, different values. I did x=Asin(wt) .035=.07sin(1.632t). .035/.07=sin(1.632t). .5/1.632=sint. t=sin^-1(.306186). Got an answer of 17.829 s which was wrong

9. Apr 24, 2010

### ehild

MissPenguins:
Sin(wt)=0.5--->wt=sin-1(0.5)= pi/6=0.5236.
w=4.375 s-1 ---> t=0.12 s.

Bearbull24.5: You can not "factor out" a multiplayer from the argument of a function. So 0.5=sin(1.632t) ---->1632t = sin-1(0.5)=pi/6.

ehild

10. Apr 24, 2010

### Bearbull24.5

So I get sin^-1(.5)=30 divide the by w (1.63299) and I get the wrong answer (18.37117)

11. Apr 24, 2010

### ehild

Write out the units. Now you got the time t in degree/rad *s. Find the angle in radians.

ehild

12. Apr 25, 2010

### MissPenguins

Can you please tell me where did you get 4.375 s from? thanks.

Nvm, I got it w = sqrt of k/m. Hehe, I got the right answer too, thank you very much.

Last edited: Apr 25, 2010
13. Apr 25, 2010

### ehild

!.375 s-1 is the angular frequency w.

w = sqrt(spring force/mass)

ehild