Question

A .5 kg mass oscillates on the end of a spring so that its horizontal position...

A .5 kg mass oscillates on the end of a spring so that its horizontal position is:

x=0.30sin(3.0t), where x is in centimeters and t is in seconds.

(I) What is the amplitude of the motion?

(II) What is the frequency, in Hertz, of the motion?

(III) What is the period of the motion?

(IV) What is the energy of the spring/mass system?

(V) What is the maximum speed of the mass?

Please explain each step. My final is tomorrow, and I don't understand this at all. Please make it as clear as possible. Thank you so much

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Homework Answers

Answer #1

(i) Amplitude, A = 0.30 cm = 0.30 * 10-2 m = 3.0 * 10-3 m

(ii) Frequency, f = ω/2π = 3.0 / 2π = 0.48 Hz

(iii) T = 2π/ω = 2π / 3.0 = 2.1 s

(iv) Total energy of the spring mass system,

TE = kA2/2 = (mω2) A2/2 = (0.5 * 3.02) * (3.0 * 10-3)2 / 2 = 2.02 * 10-5 J

(v) The total energy of the system is equal to the maximum kinetic energy of the system.

So, KEmax = TE = 2.02 * 10-5 J

=> mvmax2/2 = 2.02 * 10-5

=> vmax = [2 * (2.02 * 10-5) / 0.5]1/2 = 9.0 * 10-3 m/s

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