SHM i cant crack it!!?? (1 Viewer)

[redslert]

the question is as follows:

the period of a particle moving in SHM is 6s and its amplitude is 8cm Calculate its velocity and acceleration (1 dp) when the displacement is 5cm from the centre of the motion.

what formula do i use???

 

[ezzy85]

u know that T = 6. so use T = 2pi/n to find n. your a = 8. so sub all this into the general equation x = asin(nt + @)

with SHM, a (accelaration) = -n²x, so you can keep differentiating to get down to v, then x.

 

[redslert]

i honestly cant get it to work for some reason?!

the answer is:
v = -6.5
a = -5.5

please show me!

 

[PoLaRbEaR]

*n=pi/3 , a=8 , x=5
x = acos(nt)
v = -ansin(nt)
a = -ancos(nt) => but x = acos(nt)
a = -nx

*sub x=5 a=8 n=pi/3
5 = 8cos[(pi/3)t]
t = 49.000490121 seconds

*sub a=8 n=pi/3 t=409.000490121
v = -ansin(nt)
v = -(8pi)sin[(pi/3)49.00490121] all divided by 3
v = -6.5cm/s

*sub n=pi/3 and x=5
a = -nx
a = -(pi/3) 5
a = -5.5cm/s

 

[redslert]

Originally posted by PoLaRbEaR

x = asin(nt)
v = -ancos(nt)
[/B]

this doesnt really make sense?
if y = sinx
y' = cosx
y'' = -sinx

where did the minus come from for velocity?

also why did u use this formula: y= asin(nt+@)
how do u know which one to choose for which question???

and also THANK YOU for showing me the light, now i see it!

 

[funkychickie]

also why did u use this formula: y= asin(nt+@)
how do u know which one to choose for which question???

cause just in case it has a variable i think... it may work out that it is nothin but just in case ( i think u can think of it like a parabola, the + moves it to the left and the - moves it to the right of the origin...) someone please correct me if im wrong sorry if i confuse i really suck!!!

 

[ezzy85]

it wont matter if you use a cos or sine curve, theyre the same curve. cos is just shifted by 90 degrees.

 

[PoLaRbEaR]

ooops sorry i got the cos and sin mixed up..hold on i'll fix it..


DONE!!