Simple harmonic motion and oscillation

1. Jun 14, 2008

warmfire540

A 780-g mass is attached to a vertical spring and lowered slowly until the mass rests at its equilibrium position 30 cm below the original length of the spring. It is then set into simple harmonic motion.

(a) What is the frequency and period of the oscillation?
(b) What is the maximum velocity of the oscillating mass? Where does the maximum velocity occur?
(c) What is the maximum acceleration of the oscillating mass? Where does the maximum acceleration occur?
(d) Draw a diagram showing how the total energy of the oscillating system changes with time.

Can someone get me off to a start here? I don't understand how to find the frequency without knowing Lamda or angular frequence (w)..

2. Jun 14, 2008

Mindscrape

What are the equations that you know? We have to know what you know, so that we can pick out what you are struggling with. Maybe to get you started

$$m\ddot{x} + kx=0$$
$$x(t)=A_0 cos(\omega t + \delta)$$

where $\omega = \sqrt{k/m}$. So now you know the angular frequency.

3. Jun 15, 2008

warmfire540

Yeah, but what is k? The spring constant? or does it have another meaning?

4. Jun 18, 2008

warmfire540

I just don't get the k part..

5. Jun 18, 2008

dirk_mec1

k is the spring constant make a free body diagram and you'll see!

6. Jun 18, 2008

habibclan

Large value of k means that the spring resists being stretched, a very hard spring. A small value of k means the spring is very soft and can be stretched quite easily. Restoring force of a spring is F= -k(delta s). Opposite in direction to the displacement.

7. Jun 18, 2008

warmfire540

I still don't see.. I only know the weight, and the distance from it's equillibrium to rest position.

8. Jun 18, 2008

Mindscrape

Tell me, if you had to associate familiar physics term with weight, what would it be?

9. Jun 19, 2008

warmfire540

Huh? W=ma?

10. Jun 20, 2008

Mindscrape

Weight is a force. Are spring constants associated with a force at all? What does it mean when something is in equilibrium?

11. Jun 20, 2008

warmfire540

Oh, i see, well the weight acting down is .78*9.8=7.644 (mg)
that means there must be an equal force acting up which must be 7.644 in the opposite direction. So is this equal force the spring constant? (k)

12. Jun 20, 2008

maverick_starstrider

A force constant like k expresses the force generated per unit of some variable (whether it be displacement or acceleration)

13. Jun 21, 2008

Mindscrape

-kx=mg

in other words

14. Jun 21, 2008

warmfire540

okay, so k(-.3)=7.644
k=25.48

T=2pi*sqrt(m/k) (period of a loaded spring)
T=2pi*sqrt(.78/25.48)
T=1.099

f=1/T
f=.9096

15. Jun 22, 2008

warmfire540

just like that?

16. Jun 22, 2008

Mindscrape

Just like that.

17. Jun 22, 2008

warmfire540

Awesome!!

for part b, the max velocity is:

well, i know two equations

vmax=sqrt(k/m)A
vmax=wA (w being angular velocity)
however, i also know that velocity is greatest at the equilibrium position
so how do i find this? do i need to know k?

time out, i know k..k=25.48
so vmax=sqrt(25.48/.78)A
vmax=5.715(A) but is A (the amplitude) the 30cm?

18. Jun 22, 2008

Mindscrape

You won't be able to find a numerical answer for the amplitude with the information provided. You would need to know it's initial position and initial velocity when it was first set into harmonic motion.

An algebraic answer should be fine.

19. Jun 24, 2008

warmfire540

so just leave it at:
vmax=5.715(A)

20. Jun 26, 2008

Mindscrape

Yep, I don't see anyway around the lack of initial conditions.