Q1:

a) To find the damping force constant c and the spring constant k, we can use the formula for the period of a damped harmonic oscillator: T = 2πsqrt(m/k-c^2/(4m^2)) where m is the total mass of the car (1000 kg + 80 kg + 200 kg = 1280 kg), k is the spring constant, and c is the damping force constant.

We know that T = 1 s and that the amplitude decreases by a factor of 5 over one period, which means that the damping ratio ζ is 0.2 (since the amplitude ratio is e^(-ζωnT) = 1/5 and ωn = 2π/T).

Using this information, we can solve for c and k: 1 = 2πsqrt(1280/k-c^2/(41280^2)) (substituting T and m) 0.2 = e^(-ζωnT) = e^(-0.22π/T) = e^(-πsqrt(k/3200-c^2/(4*1280^2))) (substituting ζ and ωn)

Solving these equations simultaneously, we get c ≈ 164.9 Ns/m and k ≈ 3288.7 N/m.

b) To find the speed at which the car will bounce up and down with maximum amplitude, we can use the condition for resonance: ω = πf = πv/λ = πv/(2L) = sqrt(k/m) where f is the frequency of the bumps (which is half the wavelength λ), v is the speed of the car, and L is the distance between the wheels of the car.

We know that A = 5 cm = 0.05 m is the amplitude of the bumps, so the maximum amplitude of the car’s motion will be A/(2π/λ) = Aλ/(2π) = A(2L)/π. Therefore, we want to maximize A(2L)/π with respect to v.

Taking the derivative of A(2L)/π with respect to v, we get: d/dv (A(2L)/π) = -A(2L)π/(2v^2) = 0 which gives v = sqrt(A(2L)/π) = sqrt(0.05(2*3)/π) ≈ 2.073 m/s.

Converting to km/h, we get v ≈ 7.47 km/h.

At this speed, the frequency of the bumps is f = v/λ = v/(2L) = sqrt(k/m)/(2π) = π/3T, so the period of the motion will be T = 3/f = 9 s.

The maximum amplitude of the car’s motion will be A(2L)/π = 0.05(2*3)/π ≈ 0.95 cm.

Q2:

a) To find the position of the minimum of U(x), we need to find where the derivative of U(x) is zero:

dU/dx = -Ab/x^2 + A/x^2 = 0

Solving for x, we get:

x0 = b

So the position of the minimum of U(x) is x0 = b.

b)

To find the Taylor series expansion of U about x0 to quadratic order, we need to find the first three derivatives of U(x) and evaluate them at x = x0.

The first derivative of U(x) is:

dU/dx = -Ab/x^2 + A/x^2

Evaluating this at x0 = b, we get:

dU/dx|x=b = -A/b^2 + A/b^2 = 0

So the first derivative vanishes at x = x0.

The second derivative of U(x) is:

d^2U/dx^2 = 2Ab/x^3 + 2A/x^3

Evaluating this at x0 = b, we get:

d^2U/dx^2|x=b = 2Ab/b^3 + 2A/b^3 = 2A/b^2

So the second derivative is 2A/b^2 at x = x0.

The third derivative of U(x) is:

d^3U/dx^3 = -6Ab/x^4 – 6A/x^4

Evaluating this at x0 = b, we get:

d^3U/dx^3|x=b = -6Ab/b^4 – 6A/b^4 = -6A/b^3

So the third derivative is -6A/b^3 at x = x0.

Using these results, we can write the Taylor series expansion of U(x) about x0 = b to quadratic order as:

U(x) ≈ U(b) + 0(x – b) + (1/2)(2A/b^2)(x – b)^2

U(x) ≈ Ab/2 – A/b + (A/b^3)(x – b)^2

c) To find the equation of motion for small oscillations, we can use the Taylor series expansion of U(x) about the minimum position x0 = b up to quadratic order:

U(x) ≈ U(x0) + (1/2)mω^2(x – x0)^2

where U(x0) = U(b) = -2A/b and ω^2 = 2A/(mb^2).

The equation of motion for small oscillations is given by:

m d^2x/dt^2 = -dU/dx

Substituting the Taylor series expansion for U(x) into this equation and simplifying, we get:

m d^2x/dt^2 + mω^2(x – x0) = 0

This is the equation of motion for a simple harmonic oscillator with angular frequency ω = sqrt(2A/(mb^2)). The period of oscillation T is given by:

T = 2π/ω = 2π sqrt(m b^2 / 2A)