Question

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?

Answer 1

Answer:

(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;

$\omega _D = \sqrt{\omega_o^2 -(\frac{b}{2m})^2}\\\\\omega_o^2 = \frac{k}{m} =\frac{2.2X10^{4}}{11.3} = 1946.903rad/s\\\\thus, \omega _D = \sqrt{1946.903-(\frac{3}{2*11.3})^2} =44.12 rad/s$

The damped frequency $= \frac{\omega _D}{2\pi } = \frac{44.12}{2\pi } = 7.02 Hz$

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;

$A(t) = e^{-\frac{b}{2m}(t)}$

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

$A(t +T) = e^{-\frac{b}{2m}(t+T)}$

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

$= \frac{e^{-\frac{b}{2m}(t)}}{e^{-\frac{b}{2m}(t+T)}} =e^{-\frac{b}{2m}(T)}$

But T = 1/f

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

$=e^{-\frac{b}{2m}(T)} = e^{-\frac{3}{2X11.3}(\frac{1}{7.02})} = 0.98$

Percentage decrease = 1 - 0.98 = 0.02 = 2%