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If the mass is 48kg and the spring constant is 12, what is the period of oscillation?

a) 8 TIS
b) 4 TIS
c) TIS
d) TIS÷2s

1 Answer

3 votes

Final answer:

The period of oscillation for a mass-spring system with a 48kg mass and a spring constant of 12 is calculated using the formula T = 2π√(m/k). The correct period is 4π seconds, which is not listed in the provided options, indicating a possible error in the question.

Step-by-step explanation:

The period of oscillation for a mass-spring system undergoing simple harmonic motion can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass in kilograms, and k is the spring constant in Newtons per meter (N/m). In this case, the mass m is given as 48 kg and the spring constant k is 12 N/m.

To calculate the period, plug the given values into the formula:

T = 2π√(48/12) = 2π√(4) = 2π(2) = 4π s

Therefore, the period of oscillation is 4π seconds, which is not one of the provided options (a, b, c, or d). It appears there might be a typo or error in the options provided.

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