Question

A 55.0 g hard-boiled egg moves on the end of a spring with force constant 25.0 N/m. Its initial displacement 0.500 m. A damping force Fx=−bvx acts on the egg, and the amplitude of the motion decreases to 0.100 m in a time of 5.00 s.

Answer 1

Answer:

0.0354 kg/s

Explanation:

$A_1$ = Initial displacement = 0.5 m

$A_2$ = Final displacement = 0.1 m

m = Mass of egg = 55 g

t = Time taken = 5 seconds

Displacement of the oscillator under damping motion is given by

$x=Ae^{-\frac{b}{2m}t}cos(\omega't+\phi)$

For maximum displacement

$cos(\omega't+\phi)=1$

$A_2=A_1e^{-\frac{b}{2m}t}$

From this equation we get

$b=\frac{2m}{t}ln\frac{A_1}{A_2}\\\Rightarrow b=\frac{2\times 0.055}{5}ln\frac{0.5}{0.1}\\\Rightarrow b=0.0354\ kg/s$

The magnitude of the damping constant is 0.0354 kg/s