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.


Related Solutions

A 50.0 g object is attached to a horizontal spring with a force constant of 5.0...
A 50.0 g object is attached to a horizontal spring with a force constant of 5.0 N/m and released from rest with an amplitude of 20.0cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless? (please write out formula used)
A 45.0-g object connected to a spring with a force constant of 50.0 N/m oscillates with...
A 45.0-g object connected to a spring with a force constant of 50.0 N/m oscillates with an amplitude of 7.00 cm on a frictionless, horizontal surface. (a) Find the total energy of the system. (b) Find the speed of the object when its position is 1.30 cm. (Let 0 cm be the position of equilibrium.) (c) Find the kinetic energy when its position is 3.50 cm. (d) Find the potential energy when its position is 3.50 cm.
A 20.0-g object is placed against the free end of a spring (with spring constant k...
A 20.0-g object is placed against the free end of a spring (with spring constant k equal to 25.0 N/m) that is compressed 10.0 cm. Once released, the object slides 1.25 m across the tabletop and eventually lands 0.61 m from the edge of the table on the floor, as shown in the figure. Calculate the coefficient of friction between the table and the object. The sliding distance includes the compression of the spring and the tabletop is 1.00 m...
Take a hard boiled egg or tippee top and put it lying down on a table...
Take a hard boiled egg or tippee top and put it lying down on a table and start to spin it, if you spin it fast enough it will start to spin in an upright position. How do the egg, and tippee top remain stable in their rotating positions? What other toys can you think of are based on interesting quirks of physics? (Try to explain the physics behind them!)
A mass suspended at the end of a light Spring and spring constant K is set...
A mass suspended at the end of a light Spring and spring constant K is set into vertical motion use lagrange's equation to find the equation of motion of the mass
A mass m = 3.27 kg is attached to a spring of force constant k =...
A mass m = 3.27 kg is attached to a spring of force constant k = 60.9 N/m and set into oscillation on a horizontal frictionless surface by stretching it an amount A = 0.17 m from its equilibrium position and then releasing it. The figure below shows the oscillating mass and the particle on the associated reference circle at some time after its release. The reference circle has a radius A, and the particle traveling on the reference circle...
A spring of spring constant k sits on a frictionless horizontal surface, one end of the...
A spring of spring constant k sits on a frictionless horizontal surface, one end of the spring is attached to a wall, the other end to a block of mass M, also resting on the frictionless surface. The system executes simple harmonic motion such that at a given point the mass is -1/6 of the amplitude moving away from equilibrium, and 0.5 seconds later it has 1/7 of the maximum speeds, moving towards equilibrium, having passed equilibrium once. 1) How...
A 215 g object is attached to a spring that has a force constant of 72.5...
A 215 g object is attached to a spring that has a force constant of 72.5 N/m. The object is pulled 7.75 cm to the right of equilibrium and released from rest to slide on a horizontal, frictionless table. Calculate the maximum speed of the object. maximum speed: m/s Find the locations of the object when its velocity is one-third of the maximum speed. Treat the equilibrium position as zero, positions to the right as positive, and positions to the...
A spring with force constant k = 175 N/m is attached to the ground. On top...
A spring with force constant k = 175 N/m is attached to the ground. On top of the spring a 1.30 kg metal pan is attached. The combination could be used as a scale, but we are going to do something more interesting. We place a metal ball with mass 0.250 kg on the tray and then the tray is pushed down 0.150 m below its equilibrium point and released from rest (take this as t = 0). a) At...
The spring of a toy gun has a force constant of k = 533 N/m and...
The spring of a toy gun has a force constant of k = 533 N/m and negligible mass. The spring is compressed the length of the gun barrel, 7.25 cm, and a 0.168-g ball is placed against the compressed spring. A constant frictional force of 5.45-N acts on the ball as it travels through the barrel. The ball leaves the barrel at the moment that it loses contact with the spring. The toy gun is ‘fired’ at a height of...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT