A very bouncy mushroom can be modeled as mass of 30g(the mushroom cap) on top of a spring(the mushroom stalk) when spring constant 20 N/m. A bird of mass 50g lands on the mushroom gently so that its velocity is zero when it lands. Questions are in the image.

d) The equation that describes the oscillation of the mushroom as a function of time is; [tex]\underline{x(t) = 0.014715 \cdot cos(0.5\cdot t)}[/tex]

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]

Learn more here:

https://brainly.com/question/15487931