Question

A hard-boiled egg of mass 46.0 gg moves on the end of a spring with force constant 25.6 N/mN/m . The egg is released from rest at an initial displacement of 0.296 mm . A damping force Fx=−bvxFx=−bvx acts on the egg, and the amplitude of the motion decreases to 0.120 mm in a time of 4.55 ss .

Answer 1

Answer:

0.022kg/s

Explanation:

We are given that

Mass of boiled egg=46 g=$\frac{46}{1000} kg=0.046 kg$

$1kg=1000 g$

Constant force=F=25.6 N/m

Initial displacement=$A_1=0.296 m$

Final displacement=$A_2=0.12 m$

Time=t=4.55 s

Damping force=$F_x=-bv_x$

We have to find the  magnitude of damping constant b.

We know that the displacement of the oscillator under damping motion is given by

$x=Ae^{-\frac{b}{2m}t}cos(w't+\phi)$

For maximum displacement $cos(w't+\phi)=1$

Therefore , $x=A_2$

Substitute the values

$A_2=A_1e^{-\frac{-b}{2m}t}$

$e^{-\frac{b}{2m}t}=\frac{A_2}{A_1}$

$-\frac{b}{2m}t=ln\frac{A_2}{A_1}$

$lnx=y\implies x=e^y$

Substitute the values

$-\frac{b}{2\times 0.046}\times 4.55=ln\frac{0.12}{0.296}$

$\frac{2\times 0.046}{4.55b}=ln\frac{0.296}{0.12}$

$\frac{2\times 0.046}{4.55}=0.9b$

$b=\frac{2\times 0.46}{4.55\times 0.9}=0.022kg/s$

Hence,the  magnitude of damping constant b=0.022kg/s