Answer: 1m, 1m, 2m, 2m, 4m, 4m.
Take note that the number of oscillations are independent of their masses attached.
Explanation:
To get the number of oscillations (complete cycles) we would make use of the formula number of oscillation (n) = time (t) / period (T) ……equation 1
The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass.
period (T)= 2 x π x √(L/g) ….equation 2
where π = 3.142, L= length of the rope and g = 9.8 m/s (acceleration due to gravity)
from the question we were given time (t) = 60 secs
combining equations 1 and 2 we get
number of oscillations = time / (2 x π x √(L/g))
case 1: when L = 4m
number of oscillations = 60 / ( 2 x 3.142 x √(4/9.8))
= 14.9 = 14 complete cycles ( the question states complete cycles)
Case 2: when L = 2
number of oscillations = 60 / ( 2 x 3.142 x √(2/9.8))
= 21.4 = 21 complete cycles
Case 3: when L = 4m is the same as case 1 = 14 complete cycles
Case 4: when L = 2m is the same as case 2 = 21 complete cycles
Case 5: when L = 1m
number of oscillations = 60 / ( 2 x 3.142 x √(1/9.8))
= 30.1 = 30 complete cycles
Case 8: when L = 1m is the same as case 5 = 30 complete cycles
From the calculations above, the ranking of the pendulums from the one with the highest number of complete cycles to the lowest number of complete cycles is as follows: 1m, 2m, 2m, 4m, 4m.
Take note that the number of oscillations are independent of their masses attached.
Rank the pendulums according to the number of complete cycles of motion each pendulum goes through per minute(from high to low cycles/minute) : L = 1, L = 2, L = 4
Simple Harmonic Motion is the motion of an object (repetitive movement) through its equilibrium point where the frequency of vibrations/oscillation per second is constant
The motion that occurs in the pendulum is one example of simple harmonic motion
Terms used:
Whereas the relationship T and f:
[tex]\displaystyle T=\frac{1}{f}\\\\f=\frac{n}{t}[/tex]
n = number of vibrations
/oscillation
t = time (s)
Since the question asked is the number of complete cycles of motion each pendulum goes through per minute, then what is asked in the problem is the frequency of the pendulum
In the pendulum swing, the swing period depends on the length of the rope and gravity. The greater the length of the rope, the greater the period and the smaller the frequency
Formula used:
[tex]\large{\boxed{\bold{T=2\pi \sqrt{\frac{L}{g} }}}\rightarrow \frac{1}{f}=2\pi \sqrt{\frac{L}{g} }[/tex]
From this formula shows that the frequency is inversely proportional to the length of the rope. The longer the rope, the smaller the vibration/oscillation every minute
So that the pendulum motion from high to low frequency:
L = 1, L = 2, L = 4
(mass does not affect the pendulum swing period / frequency)
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Keywords: Simple Harmonic Motion, complete cycles, pendulum
