1A) Consider a harmonic oscillator with frequency 5.24 Hz. If the amplitude of t

Question

1A) Consider a harmonic oscillator with frequency 5.24 Hz. If the amplitude of the oscillation is 7.48 cm, what is the speed of the mass at the equilibrium point? (work this out using conservation of energy. Practise how to draw an energy bar chart for this.)

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1B) Consider a harmonic oscillator with mass 0.47 kg and spring constant 182 N/m. If the amplitude is 8.01 cm, what is the speed of the mass at a point which is displaced by 58% of the amplitude off the equilibrium point? (work this out using conservation of energy. Practise how to draw an energy bar chart for this.)

Answer with units of m/s.

Answer with exactly 3 significant figures, (for example 1.23m/s or 0.345m/s or 0.0876m/s or 12.4m/s)

Explanation / Answer

1A) Energy at end-point = Energy at equilibrium point

=> kA2/2 + 0 = 0 + mvo2/2

=> vo2 = (k/m)A2 = 2A2

=> vo = A = A(2f) = 7.48 * 10-2 * 2 * 5.24 = 2.46 m/s

1B) Energy at equilibrium point = Energy at 0.58A

=> mvo2/2 + 0 = mv2/2 + k(0.58A)2/2

=> v2 = vo2 - (k/m)(0.58A)2

=> v = [vo2 - (0.58A)2]1/2 = [2.462 - (0.58 * 7.48 * 10-2 * 2 * 5.24)2]1/2 = 2.00 m/s

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