Question

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 :/

Answer 1

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

Answer 2

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)

Answer 3

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

Answer 4

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)

Answer 5

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

Answer 6

oh my bad Its this : The angular frequency.

Answer 7

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

Answer 8

Sorry i'm confused now.

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

in letter b you got the correct
answer

Answer 13

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?

Answer 14

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

Answer 15

no f=15.5*20.56

Answer 16

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