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2 answers:
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Answer:
Explanation:
a ) angular frequency ω =
k is spring constant and m is mass attached
ω =
= 3.6515 rad / s
frequency of oscillation n = 3.6515 / (2 x 3.14)
= .5814 s⁻¹
x = .1 mcos(ωt)
= .1 mcos(3.6515t)
b ) maximum speed = ωA , A is amplitude
= 3.6515 x .1
= .36515 m /s
36.515 cm /s
maximum acceleration = ω²A
= 3.6515² x .1
= 1.333 m / s²
c ) Kinetic energy at displacement x
= 1/2 m ω²( A²-x²)
potential energy =1/2 m ω²x²
so 1/2 m ω²( A²-x²) = 1/2 m ω²x²
A²-x² = x²
2x² = A²
x = A / √2
To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore
Here,
m = mass
= Velocity (Final and initial)
First case) When the particle goes from 10m/s to 20m/s
Second case) When the particle goes from 20m/s to 30m/s
As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.