# Homework Help: Ball and tray on a Spring

1. Jun 25, 2011

### Punkyc7

A 1.2 kg horizontaal uniform tray is attached to a vertical ideal spring of force constant 195N/m and a 275 g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down 13 cm below its equilibrium point called point A and released from rest.

how long does it take the ball to leave the tray from point A?

So the time it took should equal T/4+ the time it takes to get to the spot it leaves at

T=.5464

x=Acos($\omega$t)

x is .07420
A is .13

$\omega$=sqrt(k/m)=11.4979

solving for t at dsiplacement 0 I get t=.0837

so my total time is .22. Some reason that is not the right answer and i cant figure out where i went wrong

2. Jun 25, 2011

### Staff: Mentor

How did you decide that x is .07420 ? The period, T= .5464 seconds looks okay for the tray+ball on the spring.

Where do you suppose the "launch" point is?

3. Jun 25, 2011

### Punkyc7

the launch point is at point A so -.13m

for x I did -kx=mg

x=mg/k

1.475*9.81/195

=.0742m

4. Jun 26, 2011

### Staff: Mentor

Okay, so your x is the distance above point A where the spring+tray alone are at equilibrium (spring force balances gravitational force on tray). Let's call this point B.

When the tray+ball just pass point B (on the way upwards) the tray is just beginning to experience a deceleration due to the sum of spring and gravitational forces on the tray being zero at point B. This deceleration is still much less than g , so the ball will remain in contact with the tray. It's not until the downward acceleration of the tray begins to exceed g that the ball and tray will part.

So, where will the downward acceleration of the tray be g? Hint: acceleration at g sounds like free-fall...

5. Jun 26, 2011

### Punkyc7

but if you use x as .20 you cant take arccos

6. Jun 26, 2011

### Staff: Mentor

Why would you be taking the arccos of a distance? Can't have (net) units in the arccos function argument...

Where did your 0.20 (meters?) come from?

7. Jun 26, 2011

### Punkyc7

because

x=Acos($\omega$t)

im solving for t to add it to 1/4 the period.

8. Jun 26, 2011

### Staff: Mentor

In this case x must be less than the magnitude of the harmonic motion, A. So x/A is less than one.

9. Jun 26, 2011

### Punkyc7

Yes and I did that, but my answer is wrong,so im wondering were I went wrong.

10. Jun 26, 2011

### Staff: Mentor

I'm wondering, too. Perhaps you should share your methodology and calculations so that we can puzzle it out.

11. Jun 26, 2011

### Punkyc7

(arccos(.0742/.13))/sqrt(k/m)=t=.08378

.0837+ .1366=.22sec

12. Jun 26, 2011

### Staff: Mentor

The question asks for the time from point A to the launch of the ball. You shouldn't have to include the time it takes to go from the spring-compressed position to point A.

Also, if you take the point A to be the zero reference for the motion, and if you set t=0 at that point, the displacement will correspond to a sine function rather than a cosine function.

A*sin(ω*0) = 0
A*cos(ω*0) = A

13. Jun 26, 2011

### Punkyc7

So the answer should be .189430 secs
Thank you!

14. Jun 26, 2011

### Staff: Mentor

That value looks a bit large to me. What was the calculation?

15. Jun 26, 2011

### Punkyc7

.1366+

arcsin(.0742/.13)/sqrt(195/1.75)=.05754489

if you round its .19secs

16. Jun 26, 2011

### Staff: Mentor

Your total mass value looks high; 1.2kg + 0.275kg = 1.475kg

Why are you adding 0.1366? You only want the time from point A to the launch.

17. Jun 26, 2011

### Punkyc7

because point A is at 13cm below the equilibrium spot and that is 1/4 of the period

and isnt 1.2+.275=1.475

18. Jun 26, 2011