Please help in this 5 equations!!! 1. Consider a harmonic oscillator with mass 0

Question

Please help in this 5 equations!!!

1. Consider a harmonic oscillator with mass 0.21 kg and spring constant 169 N/m. If the speed of the mass at the equilibrium point is 1.72 m/s, what is the amplitude? Convert your answer to units of cm.

2. Consider a harmonic oscillator with mass 0.22 kg and spring constant 165 N/m. If the amplitude is 8.58 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.)Answer with units of m/s.

3. Consider a harmonic oscillator with frequency 4.13 Hz. If the amplitude of the oscillation is 8 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.) Answer with units of m/s

4. Consider a harmonic oscillator with mass 0.39 kg and spring constant 180 N/m. If the amplitude is 8.54 cm, what is the speed of the mass at a point which is displaced by 51% 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

5. Consider a harmonic oscillator with period 0.07 s. If the amplitude is 8.14 cm, and at a certain time the mass is found to be moving at 1.67 m/s, what is the magnitude of the displacement from the equilibrium position (answer in units of cm). Answer with units of cm. Answer with exactly 3 significant figures

Explanation / Answer

at equilibrium particle has maximum speed

v = Aw

w = sqrt(k/m)

k = spring constant

m = mass

part 1)

w = sqrt(169/.21) = 28.638 rad/s

A = v/w = 1.72/28.638 = 0.0606 m = 6.06cm

part 2 )

at equilibrium point total energy equal to kinetic energy

1/2 * kA^2 = 1/2 * mv^2

v = sqrt(kA^2/m)

v = 2.35 m/s

part 3 )

w = 2pif

v = Aw

v = 2.076 m/s

part 4 )

1/2 mv^2 + 1/2 kx^2 = 1/2*kA^2

v = sqrt(k(A^2-x^2)/m)

x = 0.49 A

v = 1.60 m/s

part 5 )

w = 2pi/T

w = 2pi/0.07

v = w *sqrt(A^2- x^2)

v^2 = w^2 A^2 - w^2x^2

w^2x^2 = w^2A^2 - v^2

x = sqrt([w^2A^2 - v^2]/w^2)

x = 0.0792 m = 7.92 cm

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