Answer:
0.785 m/s
Explanation:
Hi!
To solve this problem we will use the equation of motion of the harmonic oscillator, i.e.
[tex]x(t) = A cos(\omega t )+B sin(\omega t)[/tex] - (1)
[tex]\frac{dx}{dt}(t) = \omega (B cos(\omega t )- A sin(\omega t)[/tex] - (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:
[tex]x(0) = 0.100[/tex]
[tex]\frac{dx}{dt}(0) = 0[/tex]
Since cos(0)=1 and sin(0) = 0:
[tex]x(0)=A[/tex]
[tex]\frac{dx}{dt}(0) = -B\omega[/tex]
We get
[tex]A =0.100\\B = 0[/tex]
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:
[tex]x(0.4) = - 0.100[/tex]
Since
[tex]x(t) = 0.100 cos(\omega t)\\ -0.100=x(0.4)=0.100cos(\omega 0.4)[/tex]
This is the same as:
[tex]-1 = cos(0.4\omega)[/tex]
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, i.e. when the sping is not stretched, that is when x = 0
With this info we will know at what time it happens:
[tex]0 = x(t) = 0.100cos(\omega t)[/tex]
The first time that the cosine is equal to zero is when its argument is equal to π/2
i.e.
[tex]t_{maxV}=\pi /(2\omega)[/tex]
And the velocity at that time is:
[tex]\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\omega t_{maxV})\\\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\pi/2)\\[/tex]
But sin(π/2) = 1.
Therefore, using eq(2):
[tex]\frac{dx}{dt}(t_{maxV} ) = 0.100*\omega = 0.100\frac{\pi}{0.400} = \pi/4[/tex]
And so:
[tex]V_{max} = \pi / 4 =0.785[/tex]
