Simple Harmonic Help (1 Viewer)

[InVinsanity]

Hey guys i got two questions

A particle is moving in SHM with acceleration d2x/dt2= -4x ms-2. If the particle starts at the origin with a velocity of 3ms-1 find the
endpoints
exact speed when particle is 1m from origin

and

The period of particle moving in SHM in 6s and its amplitude 8cm, find the equation of the displacement.

Thanks

 

[funnytomato]

d2x/dt2= -4x = -(n^2)*x , so n=2

also t=0 x=0 v=3

let the displacement x = a*cos (2t+A) (1)
then v = -a*n*sin(2t+A) (2)

sub in initial values for x, t in (1) and v , t in (2)
you get A=pi/2 , a = -3/2 which means endpoints are x=-3/2 and x=3/2

for the next part , use v^2=n^2(a^2-x^2) where n, a and x are known

 

[funnytomato]

Hey guys i got two questions


The period of particle moving in SHM in 6s and its amplitude 8cm, find the equation of the displacement.

T=2pi/n=6 , n =pi/3
a=8

 

[Drongoski]

let the displacement x = a*cos (2t+A) (1)

For this it's easier to use the simplest form: x = a sin(nt) = asin2t ==> v = dx/dt = 2acos(nt)

 

[funnytomato]

For this it's easier to use the simplest form: x = a sin(nt) = asin2t ==> v = dx/dt = 2acos(nt)

repped

 

[kooliskool]

It's actually much easier if you go to find the equation of v in terms of x, since you want that relationship to solve the question, and as you are not likely to memorise all the formulas of simple harmonic motion except the x=asinnt, (which I didn't need to at all, they can all be worked out during questions), it's better to just integrate and get it in the question:



now since ,





As
m

As for the next part,

m/s

For your second question, it's actually a very vague question, so there's no exact answer to it, it's hasn't given enough details. So you can only assume:



Where: is found when they give you the initial point it starts from, then you can find as well.

But for the purpose of the question, it's alright to just get: