Question

A 250 g mass is attached to a horizontal spring and oscillates with a frequency of 2.1 Hz. At one instant the mass is at -4.3 cm and has a horizontal velocity of 25 cm/s.

A. What is the spring constant?

B. What is the total energy of the oscillator?

C. What is the period of oscillation?

D. What is the amplitude?

E. What is the maximum speed?

Answer 1

Page No. Date m=250gm = 0.250 kg f = 2.1 Hz x = -4. 3cm, v=25cmls constant CAS Calculate spoing K=? *W=Rmf = 15 2X3.14 X 2. 1 K Jo.250 13.188 K 10.250 - 173923344 R 0.250 2 Ka 43.48 Nm (e) period of Oscillations T=- T - I = 0.476 see 2.1 IT=0.476 see

Page No Dale Amplitude oo at any point , velocity of spring COD) mass system is calculated by VCT) = = B (83–722 Vex) = K (A²x²) (2592 = 43.48 (AŽ -43)] 625 = 173,92 [ A-18:49] JA?-18.49 = 3.5936 3D Â = 35936+18:49 = २२००९ 0.250 A = 4.6989 cm. N 0.0469 m maximum speed Vmax = Aw Vmax = 0.0469 X 13.188 Vmax = 0.6185 m/s -61.85cm

Page No Date (B) Total Energy of the Oscillater E = 1 mV max E=Ź x(0.250) x (0.61857 E=0.04781 I
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