Answer:
0.785 m/s
Explanation:
Hi!
To solve this problem we will use the equation of motion of the harmonic oscillator, <em>i.e.</em>
- (1)
- (1)
The problem say us that the spring is released from rest when the spring is stretched by 0.100 m, this condition is given as:


Since cos(0)=1 and sin(0) = 0:


We get

Now it say that after 0.4s the weigth reaches zero speed. This will happen when the sping shrinks by 0.100. This condition is written as:

Since

This is the same as:

We know that cosine equals to -1 when its argument is equal to:
(2n+1)π
With n an integer
The first time should happen when n=0
Therefore:
π = 0.4ω
or
ω = π/0.4 -- (2)
Now, the maximum speed will be reached when the potential energy is zero, <em>i.e. </em>when the sping is not stretched, that is when x = 0
With this info we will know at what time it happens:

The first time that the cosine is equal to zero is when its argument is equal to π/2
<em>i.e.</em>

And the velocity at that time is:

But sin(π/2) = 1.
Therefore, using eq(2):

And so:

(i) |α| = 235.6rad.s / 0.502s = 469 rad/s²
(ii) tang a = α*r = 469rad/s² * 0.12m / 2*11 = 2.56 m/s²
Use w=m*g value of g is 1.67m/s^2