11.9 kg object oscillates at the end of a vertical spring that has a spring cons

Question

11.9 kg object oscillates at the end of a vertical spring that has a spring constant of 1.65 Times 10^4 N/m. The effect of air resistance is represented by the damping coefficient b = 3.00 N middot s/m. Calculate the frequency of the dampened oscillation. By what percentage does the amplitude of the oscillation decrease in each cycle? Find the time interval that elapses while the energy of the system drops to 4.00% of its initial value. A baby bounces up and down in her crib. Her mass is 10.0 kg, and the crib mattress can be modeled as a light spring with force constant 740 N/m. The baby soon learns to bounce with maximum amplitude and minimum effort by bending her knees at what frequency? If she were to use the mattress as a trampoline-losing contact with it for part of each cycle-what minimum amplitude of oscillation does she require?

Explanation / Answer

a)

here

by using the formula

w = sqrt( (k/m) - ( b/(2*m))^2 )

w = sqrt ( ( 1.65 * 10^4 / 11.9) - (3 / 23.8)^2)

w = 37.23 rad/s

then

w = 2 * pie * f

f = 37.23 / (2 * 3.14 )

f = 5.92 Hz

b)

the position of the mass is given by

x = A0 * e^(-bt/2m) * cos(w*t + theta)

where A0 is the amplitude without damping and A0 * e^(-bt/2m) is the amplitude with damping at a time t . After one period the amplitude becomes A0*e^(-b *(t + T)/2m and the fractional change (deltaA/A) is then

delta A /A = A0 * e^(-bt /2m) - A0*e^(-b*(t + T)/2m / ( A0 * e^(-bt / 2m)

delta A/ A = 1 - e^(-bT/2m)

since T = 1 / f

delta A / A = 1 - e^(-bT /2m)

delta A / A = 1 - e^(-b / (2mf))

= 1 - e^( -3 / (2*11.9 * 5.92 ))

= 0.021 = 2.1 %

c)

the total energy of the system is

E = 0.5 * k * A^2

= 0.5 * k * A0^2 * e^(-2bt/2m)

= E0 * e^(-bt/m)

here E0 = 0.5 * k * A0^2 is the total energy without damping . The time after which the total energy drops by 4% is then

E = 0.04 E0

E = E0 * e^(-bt/m)

0.04 = e^(-bt/m)

ln(0.04) = - 3 * t / 11.9

t = 12.76 sec

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