For a damped oscillator with a mass of 360 g, a spring constant 130 N/m and a damping coefficient of 86 g/s, what is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 11 cycles

To do this, we need to use several formulas and expressions. First, we need to know the period of time of the oscillator. This can be calculated using the following expression:

ω = 2π/T (1)

But angular frequency (ω) can be also be calculated using:

ω = √(k/m) (2)

Using (1) and (2), we can solve for the period T:

2π/T = √(k/m) (mass in kg)

2π/T = √(130/0.360)

2π/T = √361.11

2π/T = 19.003

T = 2π/19.003 = 0.331 s

Now, the amplitude A at a time t, is:

A = x exp(-bt/2m) (3)

At time 0, A = x. so A₀ = x

The problem states that we have 11 cycles respect to the initial amplitude, so expression (3) can be rewritten as:

A = x exp(-b(17t/2m)) using b as kg/s = 0.086 kg/s

Replacing the data we have:

A = x exp(-0.086(17*0.331)/2*0.36)

A = x exp(-0.086 * 7.815)

A = x exp(-0.6721)

A = 0.5106x (4)

Now, doing the ratio with the innitial we have:

A / A₀ = 0.5106x / x

The ratio is:

A/A₀ = 0.5106

Hope this helps