Answer: a) 7.5 m b) 0.45 m/s
Explanation:
a) For a simple pendulum, it can be showed that the period of the SHM that the pendulum does is as follows:
T = 2π √l/g, where l is the length of the string that is attached to the bob of the pendulum, in this case, the lamp chain.
Solving for l, we have:
l = (T/2π)² . g = (5.5)² s² / (2π)² . 9.8 m/s² = 7.5 m
b) For small angles, we can apply the conservation of energy, between the point of lowest potential energy (where speed is maximum) and the point where the potential energy is highest, and speed is zero, as follows:
U₁ + K₁ = U₂ + K₂
As we can choose freely our reference level for gravitational potential energy, we define U₁ = m. g. 0 = 0
so, at this point, K₁ = Kmax = 1/2 m vmax²
In the highest point of trajectory, v=0, ⇒ K₂ = 0
U₂ = m.g. h
Now, this height h, for small angles, can be approximated by the following expression:
h = l -l cosθ, where θ is the angle that the lamp chain aparts from vertical, in this case, 3º.
Replacing by the values, h is as follows:
h = 7.51 m - 7.51. cos 3º= 0.01 m
Now, we can write the following:
U₂ = K₁ ⇒ m.g. h = 1/2 mvmax² ⇒ vmax = √2.g.h = 0.45 m/s
The current length of the lamp is equal to 7.5m, while the maximum speed of the lamp at its maximum vertical angle is equal to 0.44m/s.
We can arrive at this answer as follows:
[tex]I=(T/2\pi)^2*g\\I= (5.5)^2s^2/(2\pi)^2*9.8m/s\\I=7.5[/tex]
[tex]U_2=2*g*h[/tex]
[tex]h= I-Icos[/tex] θ
[tex]h= 7.51-7.51cos3º\\h= 0.01m[/tex]
[tex]U_2= 2*g*h\\U=\sqrt{2(9.8*0.01)} \\U=0.44m/s[/tex]
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