Respuesta :
Answer:
a) A = 0.0221 m, b) T = 0.314 s
Explanation:
For this exercise we must separate the process into two parts, one when the body falls and another when it hits the tray
Let's look for the speed with which it reaches the tray, using energy conservation
Initial. Highest point
Em₀ = U = m g h
Final. Tray Point
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v = √2gh
Now let's apply moment conservation
Initial. Just before the crash
p₀ = m v
Final. After the crash
p_{f} = (m + M) v_{f}
p₀ = p_{f}
m v = (m + M) v_{f}
v_{f} = m / (m + M) √2gh
This is the turkey + plate system speed, which is the one that will oscillate
b) The angular velocity of the oscillation is
w = √ k / m
The angular velocity is related to the frequency and period
w = 2πf = 2π / T
T = 2π √m / k
T = 2π √ ((0.100 + 0.400) / 200)
T = 0.314 s
a) To find the amplitude let's use the equation that describes the oscillatory motion
y = A cos (wt + fi)
Speed is
v = dy / dt = - A w sin (wt + fi)
At the initial point the turkey + plate system has a maximum speed, in the previous equation the speed is maximum when cos (wt + fi) = ±1
v = v_{f} = A w
m / (m + M) √ 2gh = A w
A = m / (m + M) √2gh 1 / w
w = 2π / T
Let's calculate
A = 0.100 / (0.100 + 0.400) √(2 9.8 0.250) 0.314/2π
A = 0.2 2.2136 0.049975
A = 0.0221 m
