The motion of the bouncy mushroom can be described as a simple
harmonic motion, SHM.
Reasons:
The given parameters are;
Mass of the mushroom cap, m = 30 g = 0.03 kg
Mass of the bird = 50 g = 0.05 kg
The spring constant, K = 20 N/m
a) The equilibrium height of the mass spring system, is given as follows;
F = -K·x
[tex]x = \dfrac{F}{K}[/tex]
The applied force, F = The weight of the bird
∴ F = (0.05 kg) × 9.81 m/s² = 0.4905 N
[tex]x = \dfrac{0.05 \, kg \times 9.81 \ m/s^2}{20 \, N/m} = 0.024525 \, m[/tex]
The equilibrium height of the mushroom is 0.024525 m below its initial height.
b) The frequency of oscillation of a spring, ω, is given as follows;
[tex]\omega = \sqrt{\dfrac{K}{m} }[/tex]
Therefore;
[tex]\omega = \sqrt{\dfrac{20 \, N/m }{80 \, kg} } = \dfrac{1}{2} \, Hz[/tex]
The frequency of resulting oscillation, ω = [tex]\dfrac{1}{2} \, Hz[/tex] = 0.5 Hz
c) The applied force, F = The weight of the bird and the mushroom cap
F = (0.03 kg + 0.05 kg) × 9.81 m/s² = 0.7848 N
[tex]x = \dfrac{0.7848 \, N}{20 \, N/m } = 0.03924 \, m[/tex]
The maximum compression of the mushroom = 0.03924 m
d) The motion of the mushroom is a Simple Harmonic Motion, SHM.
The equation of a SHM as a function of time is; x(t) = A·cos(ω·t + Φ)
For the mushroom, we have;
The amplitude, A = 0.03924 m - 0.024525 m = 0.014715 m
Ф = The phase angle
When t = 0, cos(ω × 0 + Φ) = 1
cos(Φ) = 1
Ф = arcos(1) = 0
The equation is therefore;
x(t) = 0.014715·cos(0.5·t)
Equation of the oscillation of the mushroom is; [tex]\underline{x(t) = 0.014715 \cdot cos(0.5\cdot t)}[/tex]
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