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If a simple harmonic oscillator has a spring constant 200N / m

mass of 6 kg, and Total energy of 200 J. What is its maximum
velocity?

User Pabinator
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1 Answer

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Final answer:

The maximum velocity of a simple harmonic oscillator can be determined using the equation v_max = Aω, where v_max is the maximum velocity, A is the amplitude of the oscillation, and ω is the angular frequency. The angular frequency can be determined using the equation ω = sqrt(k/m), where k is the spring constant and m is the mass of the system. Substituting the values into the equations, the maximum velocity is found to be approximately 54.23 m/s.

Step-by-step explanation:

The maximum velocity of a simple harmonic oscillator can be determined using the equation:

vmax = Aω

Where vmax is the maximum velocity, A is the amplitude of the oscillation, and ω is the angular frequency.

The angular frequency can be determined using the equation:

ω = sqrt(k/m)

Where k is the spring constant and m is the mass of the system.

In this case, the spring constant is 200 N/mm, which is equivalent to 200,000 N/m, and the mass is 6 kg. Therefore,

ω = sqrt(200,000/6)

ω ≈ 271.16 rad/s

Substituting the values of A and ω into the equation for maximum velocity:

vmax = (0.200 m)(271.16 rad/s) ≈ 54.23 m/s

User Dusty
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