A thin-walled cylindrical steel water storage tank 30 ft in diameter and 62 ft long is oriented with its longitudinal axis verti
1 answer:
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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: