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An undamped 2.55 kg horizontal spring oscillator has a spring constant of 36.2 N/m. While oscillating, it is found to have a speed of 3.85 m/s as it passes through its equilibrium position.
What is its amplitude A of oscillation?
What is the oscillator's total mechanical energy E_total​ as it passes through a position that is 0.666 of the amplitude away from the equilibrium position?

An undamped 255 kg horizontal spring oscillator has a spring constant of 362 Nm While oscillating it is found to have a speed of 385 ms as it passes through its class=

Respuesta :

MathPhys

Answer:

1.02 m

18.9 J

Explanation:

The kinetic energy at the equilibrium position equals the elastic energy at the maximum displacement.

KE = EE

½ mv² = ½ kx²

mv² = kx²

x² = mv² / k

x² = (2.55 kg) (3.85 m/s)² / (36.2 N/m)

x = 1.02 m

There is no friction, so the total mechanical energy is constant at all positions. Therefore, it is equal to the kinetic energy at the equilibrium position.

E = KE

E = ½ mv²

E = ½ (2.55 kg) (3.85 m/s)²

E = 18.9 J

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