Question

A 45.0 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. Calculate the magnitude of the damping constant b. Express the magnitude of the damping coefficient numerically in kilograms per second, to three significant figures.

Answer 1

Answer:

0.02896 kg/s

Explanation:

$A_1$ = Initial displacement = 0.5 m

$A21$ = Final displacement = 0.1 m

t = Time taken = 0.5 s

m = Mass of object = 45 g

Displacement is given by

$x=Ae^{-\dfrac{b}{2m}t}cos(\omega t+\phi)$

At maximum displacement

$cos(\omega t+\phi)=1$

$\\\Rightarrow A_2=A_1e^{-\dfrac{b}{2m}t}\\\Rightarrow \dfrac{A_1}{A_2}=e^{\dfrac{b}{2m}t}\\\Rightarrow ln\dfrac{A_1}{A_2}=\dfrac{b}{2m}t\\\Rightarrow b=\dfrac{2m}{t}\times ln\dfrac{A_1}{A_2}\\\Rightarrow b=\dfrac{2\times 0.045}{5}\times ln\dfrac{0.5}{0.1}\\\Rightarrow b=0.02896\ kg/s$

The magnitude of the damping coefficient is 0.02896 kg/s