The total mechanical energy of a harmonic oscillator that consists of a spring w

Question

The total mechanical energy of a harmonic oscillator that consists of a spring with spring constant k and mass m is Etot = 1/2mv^2+1/2kx^2 Here K = 1/2mv^2 Is the kinetic energy of the mass and Uel = 1/2kx^2 is the elastic potential energy of the spring. If there are no other dissipative forces acting (e.g. friction or airdrag are negligible). the total mechanical energy is conserved. This means the oscillator reaches maximum displacement Xmax = A (amplitude) for v=0. and maximum speed for x=0 Consider a harmonic oscillator with mass 0.26 kg and spring constant 178 N/m. If the speed of the mass at the equilibrium point is 1.83 m/s what is the amplitude? (work this out using conservation of energy. Practice how to draw an energy bar chart for this ) Convert your answer to units of cm Consider a harmonic oscillator with mass 0.48 kg and spring constant 168 N/m. If the amplitude is 8.38 cm, what is the speed of the mass at the equilibrium point? (work this out using conservation of energy. Practice how to draw an energy bar chart for this.) Answer with units of m/s. Consider a harmonic oscillator with frequency 6.76 Hz If the amplitude of the oscillation is 5.54 cm, what is the speed of the mass at the equilibrium point? (work this out using conservation of energy Practice how to draw an energy bar chart for this.) Answer with units of m/s.

Explanation / Answer

Question:1

m =0.26 kg,k =178 N/m, vmax =1.83 m/s

From conservation of energy

(1/2)mvmax^2 =(1/2)kA^2

A =vmax[m/k]^1/2 = (1.83) [0.26/178]^1/2

A= 6.994 cm

Question:2

m =0.48 kg,k =168 N/m, A = 8.38 m/s

From conservation of energy

(1/2)mvmax^2 =(1/2)kA^2

vmax = A[k/m]^1/2 = (0.0838) [168/0.48]^1/2

vmax= 1.568 m/s

Question :3

f =6.76 Hz, A =5.54 cm

w=2pi*f

vmax =Aw = (0.0554)(2*3.14*6.76)

vmax= 2.352 m/s

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