A 975 g block oscillates on the end of aspring whose force constant is k = 70 N/

Question

A 975 g block oscillates on the end of aspring whose force constant is k = 70 N/m. The mass moves in a fluid which offers aresistive force F = -bv, where b = 0.162N·s/m. (a) What is the period of the motion?
1 s

(b) What is the fractional decrease in amplitude per cycle?
2

(c) Write the displacement as a function of time if at t =0, x = 0, and at t = 1.00 s, x = 0.120m.
x = ( 3 m ) exp(4s-1t ) sin ( 5 s-1t) (a) What is the period of the motion?
1 s

(b) What is the fractional decrease in amplitude per cycle?
2

(c) Write the displacement as a function of time if at t =0, x = 0, and at t = 1.00 s, x = 0.120m.
x = ( 3 m ) exp(4s-1t ) sin ( 5 s-1t)

Explanation / Answer

The mass of the block oscillating on the end of a spring is m= 975 g = 975 * 10-3 kg The spring constant of the spring is k = 70 N/m The mass moves in a fluid which offers a resistive force F =-bv, where b = 0.162 N·s/m. (a)The angular frequency of the spring is w = (k/m)1/2 or 2f = (k/m)1/2 or (2/T) = (k/m)1/2 or T = 2 * (m/k)1/2 Substituting the values in the above equation,we get T = 2 * 3.14 * (975 * 10-3/70)1/2 or T = 0.74 s (b)The fractional decrease in amplitude per cycle is F = k * x or x = (F/k) = (m * g/k) Substituting the values in the above equation,we get x = (975 * 10-3 * 9.8/70) or x = 0.1365 m (c)The displacement as a function of time is x = A * expwt * sinwt Here,w = (k/m)1/2 or w = (70/975 * 10-3)1/2 or w = 8.47 s-1 When t = 0,x = 0 or 0 = A * exp(8.47 * 0) * sin(8.47 * 0) or 0 = 0 Similarly,when t = 1.00 s,x = 0.120 m or 0.120 = A * exp(8.47 * 1.00) * sin(8.47 * 1.00) or 0.120 = A * exp(8.47) * sin(8.4) or A = (0.120/exp(8.47) * sin(8.4)) = (0.120/8.47e *0.1473) = (0.120/8.47 * 2.71 * 0.1473) or A = 0.0355 m Therefore,the displacement as a function of time is x = 0.0355 * expwt * sinwt When t = 0,x = 0 or 0 = A * exp(8.47 * 0) * sin(8.47 * 0) or 0 = 0 Similarly,when t = 1.00 s,x = 0.120 m or 0.120 = A * exp(8.47 * 1.00) * sin(8.47 * 1.00) or 0.120 = A * exp(8.47) * sin(8.4) or A = (0.120/exp(8.47) * sin(8.4)) = (0.120/8.47e *0.1473) = (0.120/8.47 * 2.71 * 0.1473) or A = 0.0355 m Therefore,the displacement as a function of time is x = 0.0355 * expwt * sinwt

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