Question

If the equation of motion of a particle is given by s = B cos(ωt + δ), the particle is said to undergo simple harmonic motion.

(a) Find the velocity of the particle at time t.

v(t) = -Bω*sin(ωt+δ)


(b) When is the velocity 0?

???


***I only need help for part (b)! thanks :)

Answer 1

velocity is zero when s= is said to be maximum
so s= max when v=0=-Bwsin(wt+d)
0=sin(wt+d)
arc sin 0=0=(wt+d)
therefore -d=wt
-d/w=t = time

Answer 2

these are the only answer choices:

http://img838.imageshack.us/img838/8430/capturebr.png

Answer 3

so the velocity is zero at t=-d/w ans

Answer 4

but that isnt one of the answer choices! :(

Answer 5

can you give us the answer choices?

Answer 6

it is linked above, in the imageshack link

Answer 7

http://img838.imageshack.us/img838/8430/capturebr.png

Answer 8

hmm i cant open the web site you given us......

Answer 9

type them out here......

Answer 10

It's the second one

Answer 11

what did you see in the site heromiles? i cant open the site ..lol

Answer 12

t = (npi - delta)/w

Answer 13

to generalized this use n instead of 2
sin[npi-(wt+d)]=0
arc sin0=0= npi-(wt+d)
wt+d=npi
wt=npi-d
t=(npi-d)/w ans

Answer 14

so the velocity is zero at t=-d/w ans or in general:
t=(npi-d)/w ans

Answer 15

hope that understand you much better..lol

Answer 16

Good job mark, but the OP is gone. He already submitted his test. So, I gave you a medal

Answer 17

ah really? hahaha that was an actual test hehehe..lol
thnx heromiles lol...have a lot of fun solving lol...

Answer 18

I explained to him how easy it was already. All he had to do was set v'(t) to zero and solve for t.

Answer 19

owkie thnx guys and enjoy..lol...