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Answer :
Final answer:
The amplitude of oscillation for the given spring-mass system is approximately 1.03 meters.
Explanation:
The question you've asked pertains to a phenomenon in physics known as Simple Harmonic Motion. In the case of an undamped oscillator like this, the maximum kinetic energy of the system occurs at the equilibrium position. Also, the total energy in an oscillating spring-mass system is constant if there's no damping, and this energy equals the potential energy of the system at its maximum amplitude. So, we can set the kinetic energy (1/2) * m * v^2 equal to the potential energy (1/2) * k * A^2, where m is the mass, v is the velocity, k is the spring constant, and A is the amplitude of oscillation. From this, we can solve for A.
The calculation is as follows: A = sqrt[((2 * m * v^2) / k)]. Plugging in the given values for mass (m = 1.28 kg), velocity (v = 2.86 m/s), and spring constant (k = 39.4 N/m), we get: A = sqrt[(2 * 1.28 kg * (2.86 m/s)^2) / 39.4 N/m] ≈ 1.03 m.
Learn more about Simple Harmonic Motion here:
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