Question

An object with mass 3.0 kg is attached to a spring with spring stiffness constant k = 300 N/m and is executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s.
(a) Calculate the amplitude of the motion.


m
(b) Calculate the maximum velocity attained by the object. [Hint: Use conservation of energy.]
m/s

Answer 1

Total energy (which is conserved)
= (1/2)M*V^2 + (1/2)k X^2
= 1.5*0.55^2 + 150*0.02^2 = 0.5138 J

(a) when V = 0 you get the maximum deflection Xmax. That is the amplitude.
0.5138 = (300/2)*Xmax^2
Xmax = 0.0585 m

(b) when X = 0 you have maximum speed:
0.5138 = 1.5*Vmax^2
Vmax = 0.5853 m/s

Answer 2

asd

Answer 3

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Answer 4

Now 12...

Answer 5

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