A body moving with simple harmonic motion has an amplitude of 1

[manal950]

A body moving with simple harmonic motion has an amplitude of 1 meter and the time for one complete oscillation is 2 seconds . Determine the velocity and acceleration of the body after 0.4 second from the mean position ?

this is the answer :
Vx = w root(r^2-x^2)
ax=w^2.X

w= 2Pi/t = 2Pi/2 = Pi rad/s
x=rcosewt
= 1 X cos Pi X 0.1 =
cose18= 0.95

Using
Vx = w root(r^2-x^2)
Vx = PI X root(1^2 - 0.95^2 )
Vx = 0.98 m/s

ax = W^2.X
=Pi^2 0.95
ax = +.37 m/s^2

My questions now form where they got t = 0.1 and and how they find X = 0.95
(the answer is not clear can please explain to me the answer )

 

[tiny-tim]

hi manal950!

(try using the X² button just above the Reply box )

0.4 s is 0.4π radians, = 72°

and the mean position is at 90°

so 72° from the mean position is at 18° (and cos18° = 0.95)