19.03.2023 - 14:58

# A 55 g rubber duck is floating in a tub of water (density 1000 k g / m 3 . The cross-sectional area of the duck is 0.0032 m 2 . The buoyancy force is given by ? A y g where buoyancy is an upward f

Question:

A 55 g rubber duck is floating in a tub of water (density 1000 {eq}kg/m^{3} {/eq}. The cross-sectional area of the duck is 0.0032 {eq}m^{2} {/eq}. The buoyancy force is given by {eq}rho Ayg {/eq} where buoyancy is an upward force generated by the water,{eq}rho {/eq} is the density of the water, A is the cross-sectional area of the floating object, y is how deep the object is submerged and g is the gravitational acceleration.

a) Find the EOM, canceling any terms that can be canceled, you may be divided y into two parts.

b) Find the duck’s natural frequency.

• April 6, 2023 в 20:16

a) The EOM (equation of motion) for the rubber duck can be derived using Newton's second law of motion which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the forces acting on the duck are gravity (weight) and buoyancy, so we can write:

Weight = mg Buoyancy = ρwater A submerged depth y g

where m is the mass of the duck, g is the gravitational acceleration, ρwater is the density of water, A is the cross-sectional area of the duck, and y is how deep the duck is submerged in the water.

Since the duck is floating, the weight and buoyancy forces are equal in magnitude, so we can set them equal to each other:

mg = ρwater A y g

Canceling out the g terms, we get:

m = ρwater A y

Solving for y, we get:

y = m/(ρwater A)

Substituting the given values, we get:

y = 55 g/(1000 kg/m^3 * 0.0032 m^2) = 0.017 m = 1.7 cm

Therefore, the duck is submerged 1.7 cm deep in the water.

b) The natural frequency of the duck can be calculated using the formula:

ω = √(k/m)

where ω is the natural frequency, k is the spring constant, and m is the mass of the duck. In this case, the buoyancy force acts as a spring, so we can calculate the spring constant using Hooke's law:

F = kx

where F is the buoyancy force, x is the displacement from the equilibrium position (i.e., the depth the duck is submerged), and k is the spring constant.

Since the buoyancy force is proportional to the submerged depth y, we can write:

F = ρwater A yg

Substituting the value of y, we get:

F = ρwater A (m/(ρwater A)) g = mg

Therefore, the spring constant is k = mg/y.

Substituting the given values, we get:

k = (55 g)(9.81 m/s^2)/(0.017 m) ≈ 3.0 N/m

Substituting the values of k and m into the formula for natural frequency, we get:

ω = √(3.0 N/m)/(55 g) ≈ 0.09 Hz

Therefore, the duck's natural frequency is \approx imately 0.09 Hz.