Question

A 50.0-g hard-boiled egg moves on the end of a spring with force constant k = 25.0 N/m. Its initial displacement is 0.300 m. A damping force

Fx=−bvx


acts on the egg, and the amplitude of the motion decreases to 0.100 m in 5.00 s. Calculate the magnitude of the damping constant b.

Answer 1

b = 0.021kg/s

Step-by-step explanation:

To find the damping constant you use the following formula, which comes from the analysis of a mass-spring constant with a damping force:

`x=x_oe^{(-b)/(2m)t}`

xo: initial amplitude = 0.300m

b: damping constant

m: mass = 0.050kg

you replace the values of the parameter and you apply properties of logarithm in oder to calculate b:

`0.100m=(0.300m)(e^{(-b)/(2(0.05kg))(5.00s)})\\\\0.333=e^(-50b)\\\\ln(0.333)=-50b\\\\b=0.021(kg)/(s)`

hence, the damping constant is 0.021kg/s

Answer 2

0.022 kg/s

Step-by-step explanation:

Given that

mass of the egg, m = 50 g = 50*10^-3 kg

Spring Force constant, k = 25 N/m

Initial displacement, A1 = 0.3 m

Final displacement, A2 = 0.1 m

Time, t = 5 s

Also, from the displacement formula used,

A = initial amplitude

b = damping constant

m = mass

t = time

w' = angular frequency

Answer is 0.022 kg/s