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A 50.0-g hard-boiled egg moves on the end of a spring with force constant k=25.0N/m. It is released with an amplitude 0.300 m. A damping force Fx=−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.

Respuesta :

Answer:

[tex]b = 21.97 gram/s[/tex]

Explanation:

As we know that amplitude of damped oscillation is given as

[tex]A = A_o e^{-bt/2m}[/tex]

so here we know that

[tex]A = 0.100 m[/tex]

[tex]A_o = 0.300 m[/tex]

t = 5.00 s

m = 50 g

now from above equation we have

[tex]0.100 = 0.300 e^{-b(5)/2(50)}[/tex]

now we have

[tex]ln(\frac{0.100}{0.300}) = -\frac{b}{20}[/tex]

[tex]b = 21.97 gram/s[/tex]

okpalawalter8

The magnitude of the damping coefficient b is mathematically given as

b = 21.97 gram/s

What is the magnitude of the damping coefficient b?

Question Parameter(s):

A 50.0-g hard-boiled egg

force constant k=25.0N/m

amplitude 0.300 m

oscillates for 5.00 s

Generally, the equation for the amplitude of damped oscillation is mathematically given as

[tex]A = A_o e^{-bt/2m}[/tex]

[tex]0.100 = 0.300 e^{-b(5)/2(50)}[/tex]

[tex]ln(\frac{0.100}{0.300}) = -\frac{b}{20}[/tex]

In conclusion  amplitude of damped oscillation is

[tex]b = 21.97 gram/s[/tex]

b = 21.97 gram/s

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