shm (1 Viewer)

[fullonoob]

The amplitude of a particle moving in shm is 5m, and its acceleration when 2m from its mean position is 4m/s^2. Find the speed of the particle at the mean position and when it is 4m from the mean position.
answer 5root2 m/s, 3 root2 m/s (how do you get this..)
i did :
x=a sin(nt+A) => when x =0, t= 0; A = 0 (lol guessed this xD)
v = an cos(nt)
v. = -an^2 sin(nt)
v. = -n^2 x
4 = - n^2 (-2)
n = +- root 2
v = -5 root2 cos(nt)
= -5root 2 m/s , t=0 at mean position
4 = 5 sin (root2 t)
t = arc sin (4/5) / root 2

help please FAILZOASJDUINADF

 

[kcqn93]

LOL, i just wasted 2 A4 pieces of paper writing gibberish...

just use v^2 = n^2(a^2 - x^2)

where n = root 2, by subbing x'' = 4 at x= 2

 

[fullonoob]

is the x in that formula x..??
and it says the resultant formula should be derived by integration each time, and not quoted, you can do the honours

 

[kcqn93]

just let x = 0 and x = 4

and for that formula i just memorize it unless they ask me to derive it. If they do ask just use v dv/dx or d(.5v^2)/dx

 

[fullonoob]

ahh ok ty