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

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

Step-by-step 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^{(-(b)/(2m))t}`

So that, the damping constant will be:

⇒
`b=(2m)/(t) log (A_(1))/(A_(2))`

On putting the estimated values, we get

⇒
`b=(2* 0.050)/(5)log (0.3)/(0.1)`

⇒
`=(0.1)/(5) log (0.3)/(0.1)`

⇒
`=0.02 \ log 3`

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