Question
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?
Answers
The period has nothing to do with the mass.
It is proportional to the square root of the length
1.97 * sqrt(2.5)
It is proportional to the square root of the length
1.97 * sqrt(2.5)
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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