The period has nothing to do with the mass.
It is proportional to the square root of the length
1.97 * sqrt(2.5)
A bob of mass m is suspended from a string of length L, forming a pendulum. The period of this pendulum is 1.97 s. If the pendulum bob is replaced with one of mass 1/2 m and the length of the pendulum is increased to 2.50 L, what is the period of oscillation?
2 answers
T1 = 1.97 s.
T2 = ?.
2pi = 6.28
T1 = 2pi*sqrt(L/g).
T1^2 = 39.5*L/g,
1.97^2 = 39.5L/9.8,
L1 = 0.962 m.
L2 = 2.50 * 0.962 = 2.41 m.
T2 = 6.28*Sqrt(2.41/9.8) = 3.11s.
T2 = ?.
2pi = 6.28
T1 = 2pi*sqrt(L/g).
T1^2 = 39.5*L/g,
1.97^2 = 39.5L/9.8,
L1 = 0.962 m.
L2 = 2.50 * 0.962 = 2.41 m.
T2 = 6.28*Sqrt(2.41/9.8) = 3.11s.
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