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:
![x(0) = 0.100](https://tex.z-dn.net/?f=x%280%29%20%3D%200.100)
![\frac{dx}{dt}(0) = 0](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%280%29%20%3D%200)
Since cos(0)=1 and sin(0) = 0:
![x(0)=A](https://tex.z-dn.net/?f=x%280%29%3DA)
![\frac{dx}{dt}(0) = -B\omega](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%280%29%20%3D%20-B%5Comega)
We get
![A =0.100\\B = 0](https://tex.z-dn.net/?f=A%20%3D0.100%5C%5CB%20%3D%200)
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:
![x(0.4) = - 0.100](https://tex.z-dn.net/?f=x%280.4%29%20%3D%20-%200.100)
Since
![x(t) = 0.100 cos(\omega t)\\ -0.100=x(0.4)=0.100cos(\omega 0.4)](https://tex.z-dn.net/?f=x%28t%29%20%3D%200.100%20cos%28%5Comega%20t%29%5C%5C%20-0.100%3Dx%280.4%29%3D0.100cos%28%5Comega%200.4%29)
This is the same as:
![-1 = cos(0.4\omega)](https://tex.z-dn.net/?f=-1%20%3D%20cos%280.4%5Comega%29)
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:
![0 = x(t) = 0.100cos(\omega t)](https://tex.z-dn.net/?f=0%20%3D%20x%28t%29%20%3D%200.100cos%28%5Comega%20t%29)
The first time that the cosine is equal to zero is when its argument is equal to π/2
<em>i.e.</em>
![t_{maxV}=\pi /(2\omega)](https://tex.z-dn.net/?f=t_%7BmaxV%7D%3D%5Cpi%20%2F%282%5Comega%29)
And the velocity at that time is:
![\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\omega t_{maxV})\\\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\pi/2)\\](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%28t_%7BmaxV%7D%20%29%20%3D-%200.100%5Comega%20sin%28%5Comega%20t_%7BmaxV%7D%29%5C%5C%5Cfrac%7Bdx%7D%7Bdt%7D%28t_%7BmaxV%7D%20%29%20%3D-%200.100%5Comega%20sin%28%5Cpi%2F2%29%5C%5C)
But sin(π/2) = 1.
Therefore, using eq(2):
![\frac{dx}{dt}(t_{maxV} ) = 0.100*\omega = 0.100\frac{\pi}{0.400} = \pi/4](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%28t_%7BmaxV%7D%20%29%20%3D%200.100%2A%5Comega%20%3D%200.100%5Cfrac%7B%5Cpi%7D%7B0.400%7D%20%3D%20%5Cpi%2F4)
And so:
![V_{max} = \pi / 4 =0.785](https://tex.z-dn.net/?f=V_%7Bmax%7D%20%3D%20%5Cpi%20%2F%204%20%3D0.785)