8. Consider the two simple pendulums with identical masses but with different le

Question

8. Consider the two simple pendulums with identical masses but with different lengths as shown below. Suppose they are released from rest from position A at the same time as shown. So you understand that they will not oscillate in harmony since they will have different periods of oscillation. But at a later time we will see both pendulums reaching the position A simultaneously. When will it take place? (calculate the time taken from the start). Take g = 980cm/s^2 . Must show all the calculations. (1.0 point)

Explanation / Answer

Given:- L1=65.2cm,L2=260.8cm

T = 2pi (L/G)^(1/2)
T1 = 2pi (0.65.2/9.80)^(1/2) = 0.80sec
T2 = 2pi(2.608/9.80)^(1/2) = 1.612sec

Alternatively, if only the first time is desired, it is easier to use this method:
the frequency of L1 is 1/T1 = 1/0.80 = 1.25Hz
and f L2 = 1/T2 = 1/1.612 = 0.617Hz
It follows that, for the first intersection,
t * f = 1
t (1.25 + 0.617) = 1
or t = 0.535sec

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