Respuesta :
Answer:
a) A' = 0.345 m, b) f = 2,800 Hz
Explanation:
b) The angular velocity of a simple harmonic motion is
w =[tex]\sqrt{\frac{k}{m} }[/tex]
angular velocity and frequency are related
w = 2π f
we substitute
f = 1 /2π √k/m
indicates that the initial frequency value f = 3.96 Hz
in this case the mass is reduced by half
m ’= m / 2
we substitute
f = 2π [tex]\sqrt{\frac{k}{m} }[/tex]
f = √1/2 (2π √k/m)
f = 1 /√2 3.96
f = 2,800 Hz
a) The amplitude of the movement is defined by the value of the initial depalzamienot before an external force that initiates the movement.
When the block is divided into two parts of equal masses as if it were exploding, for which we can use the conservation of moment
initial instant. Right before the division
p₀ = (m₁ + m₁) v
final instant. Right after the split
p_f = m₁ v '
p₀ = p_f
(2 m₁) v = m₁ v ’
v ’= 2v
At this point we can use conservation of energy for the system with only half the block.
Starting point. Where the block divides
Em₀o = K = ½ m v'²
Final point. Point of maximum elongation
Em_f = Ke = ½ k A²
how energy is conserved
Em₀ = Em_f
½ m’ v’² = ½ k A’²
we substitute the previous expressions
½ m/2 (2v)² = ½ k A’²
A’² = 2 m v² / k (1)
Let's use the conservation of energy with the initial conditions, before dividing the block
½ m v2 = ½ k A2
A² = mv² / k = 5.95 10⁻² m²
we substitute in 1
A'² = 2 A²
A ’²= 2 5.95 10⁻²
A ’²= 11.9 10⁻² m
A' = 0.345 m
