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=−bvxF



x



​



=−bv



x



​



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

Answer:

The correct answer will be "0.022 kg/s".

Explanation:

The given values are:

Initial amplitude,

A₁ = 0.3 m

Final amplitude,

A₂ = 0.1 m

Mass of a egg,

m = 50.0 g i.e., 0.050 kg

Time,

t = 5 sec

As we know,

⇒  $A_{2}=A_{1}e^{(-\frac{b}{2m})t}$

So that, the damping constant will be:

⇒  $b=\frac{2m}{t} log \frac{A_{1}}{A_{2}}$

On putting the estimated values, we get

⇒  $b=\frac{2\times 0.050}{5}log \frac{0.3}{0.1}$

⇒     $=\frac{0.1}{5} log \frac{0.3}{0.1}$

⇒     $=0.02 \ log 3$

⇒     $=0.0219 \ kg/sec \ or \ 0.022 \ kg/s$