A 11.3-kg object oscillates at the end of a vertical spring that has a spring constant of 2.20 ✕ 104 N/m. The effect of air resistance is represented by the damping coefficient b = 3.00 N · s/m.(a) Calculate the frequency of the dampened oscillation.
(b) By what percentage does the amplitude of the oscillation decrease in each cycle?

(a) the frequency of the dampened oscillation is 7.02 Hz

(b) percentage decrease in amplitude of the oscillation in each cycle is 2%

Explanation:

Given;

mass of the object = 11.3 kg

the spring constant = 2.2 X 10⁴ N/m.

damping coefficient b = 3.00 N · s/m

Part (a) the frequency of the dampened oscillation

The oscillation frequency is calculated as follows;

The damped frequency

Part (b)  percentage decrease in amplitude of the oscillation in each cycle

The amplitude of the oscillation depends on the damping coefficient (b) and period (T), and it is given as;

After one cycle, the amplitude changes from A(t) to A(t+T), where T is the period of the oscillation.

Percentage decrease in amplitude is gotten by dividing A(t) by A(t+T)

But T = 1/f

Substituting the values of the parameters in the above equation, we will have;

Percentage decrease = 1 - 0.98 = 0.02 = 2%

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.