A spring with an 8-kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equili
2 answers:
Answer:
x = 1.26 sin 3.16 t
Explanation:
Assume that the general equation of the displacement given as
x = A sinω t
A=Amplitude ,t=time ,ω=natural frequency
We know that speed V
V= A ω cosωt
Maximum velocity
V(max)= Aω
Given that F= 32 N
F = K Δ
K=Spring constant
Δ = 0.4 m
32 =0.4 K
K = 80 N/m
We know that ω²m = K
8 ω² = 80
ω = 3.16 s⁻¹
Given that V(max)= Aω = 4 m/s
3.16 A = 4
A= 1.26 m
Therefore the general equation of displacement
x = 1.26 sin 3.16 t
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Explanation:
Since, the rod is present in vertical position and the spring is unrestrained.
So, initial potential energy stored in the spring is = 0
And, initial potential gravitational potential energy of the rod is .
It is given that,
mass of the bar = 0.795 kg
g = 9.8
L = length of the rod = 0.2 m
Initial total energy T =
Now, when the rod is in horizontal position then final total energy will be as follows.
T =
where, I = moment of inertia of the rod about the end =
Also,
where, = speed of the tip of the rod
x = spring extension
The initial unstrained length is = 0.1 m
Therefore, final length will be calculated as follows.
x' = m
Then, x =
x = m - 0.1 m
= 0.1236 m
k = 25 N/m
So, according to the law of conservation of energy
Putting the given values into the above formula as follows.
v = 2.079 m/s
Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.