Question

In: Physics

A 50.0-g hard-boiled egg moves on the end of a spring with force constant k =...

A 50.0-g hard-boiled egg moves on the end of a spring with force constant k = 25.0 N/m. It is released with an amplitude 0.300 m. A damping force F_x = -bv acts on the egg. After it oscillates for 5.00 s, the amplitude of the motion has decreased to 0.100 m.

Calculate the magnitude of the damping coefficient b.
Express the magnitude of the damping coefficient numerically in kilograms per second, to three significant figures.

Solutions

Expert Solution

Concepts and reason

The concept of damped oscillation is required to solve the problem.

Initially, determine the expression for damping coefficient by using the expression of amplitude of oscillation at time t. Later, calculate the damping coefficient by using the expression of damping coefficient.

Fundamentals

Damped oscillations are the oscillations for which the amplitude of oscillation decreases with time. The amplitude of oscillation of an object at time t is given as,

A=A0ebt2mA = {A_0}{e^{\frac{{ - bt}}{{2m}}}}

Here, A0{A_0} is the amplitude at time t=0t = 0 , b is the damping coefficient, and m is the mass of the object.

The amplitude of oscillation at time t is given as,

A=A0ebt2mA = {A_0}{e^{\frac{{ - bt}}{{2m}}}}

Rearrange the above equation for b.

A=A0ebt2mebt2m=A0Abt2m=lnA0Ab=(2mt)lnA0A\begin{array}{c}\\A = {A_0}{e^{\frac{{ - bt}}{{2m}}}}\\\\{e^{\frac{{bt}}{{2m}}}} = \frac{{{A_0}}}{A}\\\\\frac{{bt}}{{2m}} = {\ln ^{\frac{{{A_0}}}{A}}}\\\\b = \left( {\frac{{2m}}{t}} \right){\ln ^{\frac{{{A_0}}}{A}}}\\\end{array}

The damping coefficient is calculated by using the relation,

b=(2mt)lnA0Ab = \left( {\frac{{2m}}{t}} \right){\ln ^{\frac{{{A_0}}}{A}}}

Substitute 50.0 g for m, 5.00 s for t, 0.300 m for A0{A_0} , and 0.100 m for A in the equation b=(2mt)lnA0Ab = \left( {\frac{{2m}}{t}} \right){\ln ^{\frac{{{A_0}}}{A}}} and solve for the damping coefficient.

b=(2(50.0g(103kg1g))5.00s)ln0.300m0.100m=0.0220kg/s\begin{array}{c}\\b = \left( {\frac{{2\left( {50.0{\rm{ g}}\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)} \right)}}{{5.00{\rm{ s}}}}} \right){\ln ^{\frac{{0.300{\rm{ m}}}}{{0.100{\rm{ m}}}}}}\\\\ = 0.0220{\rm{ kg/s}}\\\end{array}

Ans:

The magnitude of damping coefficient is 0.0220 kg/s.


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