## MarcLeclair 2 years ago A 240g particle oscillating in SHM travels 31cm between the two extreme points in its motion with an average speed of 90cm/s. b) The maximum force on the particle. Can't figure that one out :/

1. Natan

(b) In this case the force is equal to F=-kx, |F|=kx. If k =const. the maximum force is when x is maximum.

2. BoneFreeze

Natan is right k=mg/h and since Fmax = kh = mg/h *h = mg =0.24*Acceleration due gravite (9.8 m/s2 or something like that)

3. Natan

yes. that is right. please can you tell me about the letter a .

4. MarcLeclair

That doesnt work though I tried, in letter A i got the answer right I did t=1/f therefore t=2pi/w which is in turn w=9.26. Knowing the mass is .240 kg you can find out k being 20.56 no? and I get fmax = 15.5cm * 25.56 (15.5cm being the Xm)

5. Natan

that is fmax=20.56*15.5. what is the question in letter a?

6. MarcLeclair

oh my bad Its this : The angular frequency.

7. MarcLeclair

find the angular frequence* which i found with the t=1/f

8. Natan

Sorry i'm confused now.

9. MarcLeclair

In a i have to find the angular frequency, the w of the following equation x=Asin(wt+ alpha) alpha being the phase constant.

10. MarcLeclair

I found my W using the fact that frequency is T ( period) = 1/f and therefore T=2pi/w

11. Natan

it is OK. Now I urderstood. k=w^2*m

12. Natan

in letter b you got the correct answer

13. MarcLeclair

yeah so k = 9.26^2 * .240g which gives me 20.56 but it always gives it wrong to me :/ could it be the sign?

14. MarcLeclair

so f=kx therefore my amplitude is 31/2 so f = 20.56 *.240 no?

15. Natan

no f=15.5*20.56

16. MarcLeclair

yeah sorry and it g ives 319 but it says its wrong, I don't know what to do to change it