1) Consider a harmonic oscillator with mass 0.48 kg and spring constant 187 N/m.

Question

1)

Consider a harmonic oscillator with mass 0.48 kg and spring constant 187 N/m. If the amplitude is 5.26 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.)

2)

Consider a harmonic oscillator with frequency 5 Hz. If the amplitude of the oscillation is 5.73 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.)

Explanation / Answer

use the principle of conservtion of energy as KE = EPE

0.5 mV^2 = 0.5 kA^2

V is velocity

k is spring constant

A is amplitude

so here V^2 = 187 * 0.0526* 0.0526/(0.48)

V = 1.038 m/s

------------------------------------------------------

as V = Lf

where L is waveength= 2L = 2* 0.0573 m

so speed V = 2* 0.0573 * 5

V = 0.573 m/s

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