Six pendulums of mass m and length L as shown are released from rest at the same angle (theta) from vertical. Rank the pendulums

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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

Physics

2 answers:

ValentinkaMS [3.4K] · Answer 1 · 2026-03-03T00:29:43+03:00

List the pendulums based on the frequency of complete cycles each performs per minute, arranged from highest to lowest frequency: L = 1, L = 2, L = 4

Additional details

Simple Harmonic Motion refers to the repetitive motion of an object through its equilibrium position, characterized by a consistent rate of vibrations or oscillations per second.

The swinging of a pendulum exemplifies simple harmonic motion.

Definitions:

Deviation (y): the distance of the object from its point of equilibrium.
Amplitude (A): the maximum distance from the equilibrium position.
Frequency (f): the count of oscillations within a given timeframe.
Period (T): the time taken for one complete cycle.

The following relationship exists between T and f:

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

n = number of oscillations

t = time (s)

The problem inquires about the frequency of pendulum motion, which corresponds to the number of complete cycles executed each minute.

The swing period of a pendulum is influenced by the rope length and gravitational force. A longer rope results in a longer period and a lower frequency.

The formula applied is as follows:

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

This formula indicates that frequency varies inversely with rope length—greater length yields fewer oscillations per minute.

Consequently, the pendulum's motion, arranged from highest to lowest frequency, is as follows:

L = 1, L = 2, L = 4

(the mass of the pendulum does not influence the swing period or frequency)

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Keywords: Simple Harmonic Motion, complete cycles, pendulum

Sav [3.1K] · Answer 2 · 2026-03-03T00:32:55+03:00

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.