PARTA k = 2.000 N/m F = -kX M=0.125kg wood E 4 x(m) a 0.04 0.02 A 0.125kg mass is attached to a horizontal spring of spr

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Parta K 2 000 N M F Kx M 0 125kg Wood E 4 X M A 0 04 0 02 A 0 125kg Mass Is Attached To A Horizontal Spring Of Spr 1 (69.5 KiB) Viewed 24 times

PARTA k = 2.000 N/m F = -kX M=0.125kg wood E 4 x(m) a 0.04 0.02 A 0.125kg mass is attached to a horizontal spring of spring constant k = 2 N/m and set into oscillation. Assume frictionless floor. What is the angular frequency of the vibration? 6 = - 4 radis How about the vibrating frequency fin Hz? f = -0.6306 = 2 The position of the mass is given as x = A cos (ot + ). With your calculated value of o, fill in the following table don't forget to set your calculator to "radian mode"). Each row represents a different oscillation with the same spring, pay attention to sign: A (m) v(m/s) a (m/s) t(s) 00 (rad) F(N) 0.03 En Diaz D00920030169 2.5 0.3 10.0384 0.1385 0.3199 0.2118 0.2 -0.04 0.06 D-05136 0.1 1-0.8672 1.36 0.4 0.04 0.3 1-0.64 1-0.08 -0.05 -0.25 0.1 X = AcOS (wt + Do). V=-hwsin (with PARTB Each of the rows in the following table represents an oscillation of a different spring and mass. Fill in the blanks. E is the total energy. Vmax, V, Xmax, and x are all positive in values. A (m) m (kg) k (N/m) E (1) xat t=0 vat t=0 (m/s) (m/s) (rad/s) (m) (m/s) 0.05 0.25 1.25 028 0.01 1.0 9.0 0.3 0.5 0.15 2.0 5.0 -0.2 0 0 0.1 0.5 0.04 a=-A W² Coslwttbo) Vmax Xmax 20 5 2.0 ก 5.0 0.6 2.0 5.0 0.6 0.1 0.2 ง เก 2.0 1 of 2