Question

Dr C #1 An object with mass = 5 kg is oscillating on the end of a horizontal spring with spring constant 10 ^4 N/M. there is no friction, but there is air resistance, which can be represented with a damping coefficient b = 10 Ns/m. Be careful to avoid rounding errors.

A) Is the mass underdamped, overdamped, or critically damped?

B) What is the frequency of oscillation of this mass? The period?

C) What percent of energy is lost during one cycle of motion?

D) how many cycles of motion will occur in the time it takes the energy to drop to 10% of its initial value?

E) What is the quality factor of this oscillator?

Answer 1

A) b² - 4mk = (10²) - ( 4*5*10e4)

clearly b² - 4mk < 0, hence it is underdamped oscillation (ans)

B) frequency = w₀ = sqrt(k/m) = sqrt( 10^4/5) = 44.72 Hz (ans)

C) energy= 1/2mA²w₀²e^yt[(1+((y/2w₀)cos(2w_dt+x))],

y= b/2m = 1

and w_{0 =} 44.72 hz, we can neglect the cosine term in the energy

Therefore Energy= 1/2mA²w₀²e^yt

when one oscillation is over T= 2*pi/w₀ = 0.1405 seconds

percent of energy = e^yT = e^1*0.1405 = 86.9 % of intitial energy ( as m, A, w₀, is constant and initial time t= 0seconds)

,therefor percent of energy lost = 100-86.9 = 13.1 % (ans)

D) given that percent of energy left is 10 % of initial,

0.1=e^yT

as y=1 we get T=2.302 seconds

no. of cycles = 2.303/0.1405 = 16.38 cycles

E) quality factor = w_d/ y

                       = w₀/ y ( since w_d = w₀ as y is very small compared to w₀)

                       = 44.72 (ans)