A mass of 2.61 kg is connected to one end of a horizontal spring; the other end

Question

A mass of 2.61 kg is connected to one end of a horizontal spring; the other end is fixed. The mass is pulled, then released into oscillating motion. A technician measures the motion, and finds an amplitude of 0.921 m and a duration of 129 s for 79 cycles of oscillation. Find the frequency, f, the speed at the equilibrium position, max, the spring constant, k, the potential energy at an endpoint, Umax, the potential energy when the mass is located 48.1% of the amplitude away from the equilibrium position, U and the kinetic energy, K, and the speed, v, at the same position Number f- Hz Number m/ S Number N/m Number (Scroll down for more answer blanks.) Number U= Number K= Number m/s

Explanation / Answer

from the given data

A = 0.921 m

T = 129/79 = 1.633 s

a) f = 1/T = 1/1.633 = 0.612 s

b) Vmax = A*w

= A*2*pi*f

= 0.921*2*pi*0.612

= 3.54 m/s

c) we know, w = sqrt(k/m)

w^2 = k/m

(2*pi*f)^2 = k/m

==> k = (2*pi*f)^2*m

= (2*pi*0.612)^2*2.61

= 38.6 N/m

d) Umax = (1/2)*k*A^2

= (1/2)*38.6*0.921^2

= 16.4 J

e) U = (1/2)*k*x^2

= (1/2)*38.6*(0.481*0.921)^2

= 3.8 J

f) KE = 16.4 - 3.8

= 12.6 J

g) v = sqrt(2*KE/m)

= sqrt(2*12.6/2.61)

= 3.11 m/s

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