A voltage of 12 cos(I000t+45) Vis applied to a circuit in which a resistor of 4 .n, aninductor of L H, and a capacitor of 100 μF
1 answer:
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Answer:
D/H =15
Explanation:
- We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
- At this point, we can find the value of H, applying the following kinematic equation:
- If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:
- We can use the same equation, to find the value of D, as follows:
- In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
- When we replace these values in (1), we find that v₁ = -v₀.
- Replacing in (3), we have:
- Solving for D:
- From (2) we know that H can be expressed as follows:
- ⇒ D = 15 * H
Answer:
The motion is over-damped when λ^2 - w^2 > 0 or when > 0.86
The motion is critically when λ^2 - w^2 = 0 or when = 0.86
The motion is under-damped when λ^2 - w^2 < 0 or when < 0.86
Explanation:
Using the newton second law
k is the spring constante
b positive damping constant
m mass attached
x(t) is the displacement from the equilibrium position
Converting units of weights in units of mass (equation of motion)
From hook's law we can calculate the spring constant k
If we put m and k into the DE, we get
Denoting the constants
2λ = =
λ = b/0.215
λ^2 - w^2 =
This way,
The motion is over-damped when λ^2 - w^2 > 0 or when > 0.86
The motion is critically when λ^2 - w^2 = 0 or when = 0.86
The motion is under-damped when λ^2 - w^2 < 0 or when < 0.86