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A 40.0-N force stretches a vertical spring 0.250 m.

(a) What mass must be suspended form the spring so that the system will oscillate with a period of 1.0 s?

User Mikewaters
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Final answer:

To determine the mass needed for the spring to oscillate with a period of 1.0 s, we can use the formula for the period of an oscillating mass-spring system. By plugging in the given values, we can solve for the mass.

Step-by-step explanation:

To find the mass that must be suspended from the spring to achieve a period of 1.0 s, we can use the formula for the period of an oscillating mass-spring system:

Calculating the Required Mass for a Specific Oscillation Period

To calculate the mass that must be suspended from a spring to achieve a certain period of oscillation, we use the formula for the period (T) of a simple harmonic oscillator, T=2π√(m/k), where 'm' is the mass and 'k' is the spring constant. Given a spring constant (k) of 40.0 N/m and a desired period (T) of 1.0 s, we rearrange the formula to solve for mass (m): m = √((T/2π)² ∙ k). Inserting the values, we get m = √((1.0/2π)² ∙ 40.0 N/m), which simplifies to m = 0.159 kg or 159 g.

Period = 2π√(m/k)

Given that the force constant (k) of the spring is 40.0 N/m and the desired period is 1.0 s, we can rearrange the formula to solve for the mass (m):

m = (Period^2 * k) / (4π^2)

Substituting the given values, we get: m = (1.0^2 * 40.0) / (4π^2)

Simplifying the expression gives us the mass required for the system to oscillate with a period of 1.0 s.

User Xrisk
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