Physics High School

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.

Answers

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,

⇒  [tex]A_{2}=A_{1}e^{(-\frac{b}{2m})t}[/tex]

So that, the damping constant will be:

⇒  [tex]b=\frac{2m}{t} log \frac{A_{1}}{A_{2}}[/tex]

On putting the estimated values, we get

⇒  [tex]b=\frac{2\times 0.050}{5}log \frac{0.3}{0.1}[/tex]

⇒     [tex]=\frac{0.1}{5} log \frac{0.3}{0.1}[/tex]

⇒     [tex]=0.02 \ log 3[/tex]

⇒     [tex]=0.0219 \ kg/sec \ or \ 0.022 \ kg/s[/tex]

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