A vertical spring of spring constant 145 N/m supports amass of 75 g. The mass os
ID: 1727733 • Letter: A
Question
A vertical spring of spring constant 145 N/m supports amass of 75 g. The mass oscillates in a tube of liquid. If themass is initially given an amplitude of 5.5 cm, the mass isobserved to have an amplitude of 2.2 cm after4.0 s. Estimate the damping constant b. Neglect buoyant forces. b=_____ kg/s I tried using x=Ae-tcos't and =b/2m and'=((k/m)-(b2/4m2)) and got an answer of 0.074 kg/s which is wrong. I'm notsure where the problem is. A vertical spring of spring constant 145 N/m supports amass of 75 g. The mass oscillates in a tube of liquid. If themass is initially given an amplitude of 5.5 cm, the mass isobserved to have an amplitude of 2.2 cm after4.0 s. Estimate the damping constant b. Neglect buoyant forces. b=_____ kg/s I tried using x=Ae-tcos't and =b/2m and'=((k/m)-(b2/4m2)) and got an answer of 0.074 kg/s which is wrong. I'm notsure where the problem is.Explanation / Answer
. k = 145 N/m ; m = 0.075 kg ; Elapsed time (t) = 4.0 s; . Amplitude ration (A) = 0.022/0.055 =0.4 .Frequency = (k/m) = (145/0.075) = 43.97 rad/s . Period T = 2/ = ( 2*3.14 ) / 43.97 = 0.143 s ; . N = t / T = 4.0 / 0.143 = 27.972 . We know that : .
= ln(A)/N = ln (0.4) / 27.972 = -0.0327 /cycle .
= -/sqrt(42+2) = 0.0327/ ( 4 *(3.14)2 + (- 0.0327) ) = 5.20 * 10-3 . b = 2m = 2k/ =2/sqrt(km) = ( 2 * 5.20 * 10-3 ) / sqrt ( 145 * 0.075) . = --------------- . Solve it . Hope this helps u!