Answer: 1m, 1m, 2m, 2m, 4m, 4m.
It’s important to remember that the masses attached do not influence the number of oscillations.
Explanation:
To determine the number of oscillations (complete cycles), we can apply the formula n = t / T ……equation 1
The variables that impact the period of a simple pendulum are solely its length and gravitational acceleration. The period remains unaffected by factors such as mass.
period (T)= 2 x π x √(L/g) ….equation 2
where π = 3.142, L= rope length, and g = 9.8 m/s (gravitational acceleration)
According to the question, the time (t) is 60 seconds.
By merging equations 1 and 2, we obtain
number of oscillations = time / (2 x π x √(L/g))
Case 1: for L = 4m
number of oscillations = 60 / ( 2 x 3.142 x √(4/9.8))
= 14.9 = 14 complete cycles (the problem specifies complete cycles)
Case 2: for L = 2m
number of oscillations = 60 / ( 2 x 3.142 x √(2/9.8))
= 21.4 = 21 complete cycles
Case 3: where L = 4m, results in the same as case 1, yielding 14 complete cycles
Case 4: where L = 2m, mirrors the outcome in case 2, producing 21 complete cycles
Case 5: in the instance of L = 1m
number of oscillations = 60 / ( 2 x 3.142 x √(1/9.8))
= 30.1 = 30 complete cycles
Case 6: when L = 1m, which repeats case 5, also gives 30 complete cycles
From these findings, the order of the pendulums from the highest to lowest number of complete cycles is as follows: 1m, 2m, 2m, 4m, 4m.
Remember, the number of oscillations is independent of their respective masses.
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