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If the spring has a mass M and mass m is suspended from it, effective mass is given by meff = m + M/3. If two masses m1 and m2 are connected by a spring and made to oscillate on a horizontal surface, reduced mass = 1/mr = 1/m1 + 1/m2. Time period T = 2π√(mr/k).

1. T = 2π√(M/K)
2. T1/T2 = √(M1/M2)
3. T2 = T1√(M2/M1) = T1√(2M/M)
4. T2 = T1√2 = √2T (Where T1 = T)

Correct answer: T2 = T1√(M2/M1) = T1√(2M/M)

Solution

The time period for oscillation depends on the reduced mass and spring constant. Using the given formula, T2 = T1√(M2/M1) is correct as it aligns with the relationship between time period and mass.

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