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0.540 kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to change the period to 2.05 s

User Champagne
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Answer: Approximately 0.993 kg of mass must be added to the system to change the period of oscillation from 1.50 s to 2.05 s.

Step-by-step explanation:

To determine how much mass must be added to change the period of oscillation from 1.50 s to 2.05 s, we can use the formula for the period of a mass-spring system:

T = 2π√(m/k)

Where T is the period, m is the mass, and k is the spring constant.

Let's first calculate the initial spring constant using the given period of 1.50 s and the initial mass of 0.540 kg. Rearranging the formula, we have:

k = (4π²m) / T²

k = (4π² * 0.540 kg) / (1.50 s)²

Next, we can calculate the new mass that needs to be added to change the period to 2.05 s. Rearranging the formula again, we have:

m₂ = k(T₂²) / (4π²)

Where m₂ is the additional mass and T₂ is the desired period of 2.05 s.

m₂ = k * (2.05 s)² / (4π²)

Now, we can substitute the initial values and calculate the final mass:

m₂ = [(4π² * 0.540 kg) / (1.50 s)²] * (2.05 s)² / (4π²)

Simplifying the expression, the π² terms and s² terms cancel out, leaving:

m₂ = (0.540 kg) * (2.05 s)² / (1.50 s)²

m₂ = (0.540 kg) * (2.05)² / (1.50)²

m₂ ≈ 0.993 kg

Therefore, approximately 0.993 kg of mass must be added to the system to change the period of oscillation from 1.50 s to 2.05 s.

I hope this helps :)

User Robert Alexander
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