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Answer :
The maximum velocity, Vmax (in m/s) of a simple harmonic oscillator with a mass of 8 kg, a spring constant 75 N/m, and Total energy of 135 J can be determined using the formula: Vmax = √(2E/m) where E is the Total energy and m is the mass.
Substituting the given values into the formula, we get: Vmax = √(2E/m) = √(2 x 135 J/8 kg) = √(33.75)≈ 5.81 m/s Therefore, the maximum velocity of the simple harmonic oscillator is 5.81 m/s. Here is a detailed explanation for your understanding:
Given,Mass of the oscillator, m = 8 kgSpring constant, k = 75 N/mTotal energy, E = 135 J We know that the total energy of a simple harmonic oscillator is given by:E = 1/2 kx² + 1/2 mv²where x is the amplitude and v is the velocity of the oscillator.Substituting the given values, we get:135 J = 1/2 × 75 N/m × x² + 1/2 × 8 kg × v²The maximum velocity occurs at the equilibrium position, where x = 0.Therefore, the equation becomes:135 J = 1/2 × 8 kg × v²Simplifying and solving for v, we get:v = √(2E/m) = √(2 x 135 J/8 kg) = √(33.75)≈ 5.81 m/s
Therefore, the maximum velocity of the simple harmonic oscillator is 5.81 m/s.
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