Question

Six pendulums of mass m and length L as shown are released from rest at the same angle (theta) from vertical. Rank the pendulums according to the number of complete cycles of motion each pendulum goes through per minute.
L = 4 m, M = 1 kg
L = 2 m, M = 4 kg
L = 4 m, M = 4 kg
L = 2 m, M = 2 kg
L = 1 m, M = 4 kg
L = 1 m, M = 2 kg

Answer 1

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

Further explanation

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:

• Deviation (y): the object's distance from the balance point
• Amplitude (A): maximum deviation
• frequency (f): number of vibrations at any time
• Period (T): the amount of time in one vibration

Whereas the relationship T and f:

$\displaystyle T=\frac{1}{f}\\\\f=\frac{n}{t}$

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:

$\large{\boxed{\bold{T=2\pi \sqrt{\frac{L}{g} }}}\rightarrow \frac{1}{f}=2\pi \sqrt{\frac{L}{g} }$

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

Answer 2

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.