Pendulums, pendulums everywhere

Classical Mechanics Level 2

A simple pendulum consists of a small object dangling from a piece of string. A creative student constructs 8 pendulums of different lengths: $5^2 \text{ cm},\ 6^2 \text{ cm},\ 7^2 \text{ cm},\ ...,\ 11^2 \text{ cm},\ 12^2 \text{ cm},$ and hangs them next to each other in increasing order as shown in the diagram below. Thus, $L_n = \SI{(n+4)^2}{cm}$ : the shortest pendulum has length $L_1 = \SI{25}{cm},$ and the longest $L_8 = \SI{144}{cm}$ .

The student now starts the pendulums, by pulling them all sideways over the same short distance and releasing them at the same time $t = 0$ . Because of their different lengths, the pendulums swing at different frequencies and will soon all be completely out of phase.

How long will it take for all these pendulums to be exactly in phase again?

Details and Assumptions:

$g = \SI{9.81}{m/s^2}.$
The pendulums are ideal, simple pendulums with period $T_n = 2\pi\sqrt{\frac {L_n}g}.$
For a demonstration of this kind of setup, see this video .

approximately

3

minutes approximately

1\tfrac12

hours approximately

4\tfrac12

days approximately

2

weeks

1 solution

Arjen Vreugdenhil
Aug 12, 2017

The period of the pendulums are $T_n = \left(2\pi\sqrt{\frac{0.01}{9.81}}\right)(n + 4) = 0.2006(n + 4).$ They will all be in phase at a time $t$ which is an integer multiple of all $T_n$ . Thus $t = 0.2006\ N,$ where $N$ is a common multiple of $5, 6, 7, 8, 9, 10, 11, 12$ . Now $\text{lcm}(5,6,7,8,9,10,11,12) = 27\,720;$ so that the first time the pendulums are in phase again will be $t = 0.2006\cdot 27\,720 = \SI{5561}{s},$ or approximately $\boxed{1\tfrac12\ \text{hour}}$ .

Where does the 0.01 in the numerator come from?

Imi Borbas - 3 years, 9 months ago

I see now, it's to convert centimetres to metres.

Imi Borbas - 3 years, 9 months ago

In this case, the pendulums will line up when they are back in their initial position, i.e. after completing an integral number of periods. This is not always true! See this problem for an illustration.

Arjen Vreugdenhil - 3 years, 10 months ago

how did you get that formula?

Vinayak Hegde - 3 years, 9 months ago

The formula for period of a pendulum is $2\pi\sqrt{\ell/g}$ . This was stated in the problem. Now $\ell = L_n = \SI{(n+4)^2}{cm} = \SI{0.01\cdot(n+4)^2}{m},$ and we can bring the $(n+4)$ outside of the square root.

I see now that some edited the explicit statement $L_n = \SI{(n+4)^2}{cm}$ out of my problem. I will put it back in.

Arjen Vreugdenhil - 3 years, 9 months ago

Why do you find the lcm of 5, 6, 7, etc. rather than 25, 36, 49...?

Matt H - 3 years, 9 months ago

Because the period is proportional to $\sqrt L$ .

Arjen Vreugdenhil - 3 years, 9 months ago

How can you assume the pendulums will all be in phase at a time $t$ which is an integer multiple of all $T_n$ so that the first time the pendulums are in phase again will be:

$t = 0.2006\cdot 27\,720 = \SI{5561}{s},$

Although the result will be the same, but the pendulums will be meet in the same phase if it reaches the "Least Common Multiple" value of each period, not as a result of multiplier of each period. Because not necessarily the multiplication product of each period will be equal to the "Least Common Multiple" value of each period.

Alimun Bidati Sudur - 3 years, 9 months ago

The least common multiple is, by definition , the smallest positive value $t$ that is a multiple of all $T_n$ . To find it, I first wrote $T_n = N_n\:\tau,$ where all $N_n$ are integers and $\tau \approx \SI{0.2006}{s}$ . The solution is then $t = N\tau = \text{lcm}(N_5,\dots,N_{12})\:\tau.$

To find the least common multiple, I did not multiply all periods (which would have given the meaningless value $\SI{52.35}{s^8}$ ) nor did I multiply all $N_n$ values (which would have resulted in $N = 19\,958\,400$ ).

Instead, $\text{lcm}(N_5,\dots,N_12) = \text{lcm}(5, 2\cdot 3, 7, 2^3, 3^2, 2\cdot 5, 11, 2^2\cdot 3) \\ = 2^3\cdot 3^2\cdot 5 \cdot 7 \cdot 11 = 27\,720.$ That this is the smallest integer multiple of all eight periods can be shown explicitly: $t = 5544\:T_5 = 4620\:T_6 = 3960\:T_7 = 3465\:T_8 \\ = 3080\:T_9 = 2772\:T_{10} = 2520\:T_{11} = 2310\:T_{12}.$ These eight coefficients have no common factors, and are therefore the lowest integer values for which the pendulums are aligned.

Arjen Vreugdenhil - 3 years, 9 months ago

I just regret your paradigm of period multiplication (your statement: "the pendulums will all be in phase at a time $t$ which is an integer multiple of all: $T_n$ "), but in my perspective the most right is the "Least Common Multiple" value of each period for the pendulums will all be in the same phase.

Where as: An integer multiple of all $T_n$ , has not always the same value as the "Least Common Multiple ".

Alimun Bidati Sudur - 3 years, 9 months ago

@Alimun Bidati Sudur – To be precise:

The pendulums will be in phase whenever $t$ is an integer multiple of all $T_n$ (a common multiple ).
The pendulums will be in phase for the first time when $t$ is the least common multiple .

Note also that saying "an integer multiple of all $T_n$ " does not mean "multiply all $T_n$ ". Multiplying all the periods would be meaningless.

Arjen Vreugdenhil - 3 years, 9 months ago

@Arjen Vreugdenhil – Great conclusion, that's the point :)

Alimun Bidati Sudur - 3 years, 9 months ago

@Arjen Vreugdenhil – Why is the duration 27720 x 0,2006? Shouldn't it be equal to 27720 x the smallest period, I.E. 5th more ? The smallest is T5 around 1 second.

Thierry Adloff - 3 years, 9 months ago

@Thierry Adloff – No, it is 27720 times the greatest common divisor of the periods.

Arjen Vreugdenhil - 3 years, 9 months ago

@Arjen Vreugdenhil – For sure your are right. Thanks.

Thierry Adloff - 3 years, 9 months ago

# -*- coding: utf-8 -*-
"""
Created on Wed Nov 29 18:20:37 2017

@author: Fitzgeralds
"""

#https://brilliant.org/weekly-problems/2017-08-21/intermediate/?p=2

# A simple pendulum consists of a small object dangling from a piece of string.
# A creative student constructs 8 pendulums of different lengths: 
# The student now starts the pendulums, by pulling them all sideways over the same 
# short distance and releasing them at the same time. 
#Because of their different lengths, the pendulums swing at different frequencies 
#and will soon all be completely out of phase.

#How long will it take for all these pendulums to be exactly in phase again?    

from math import pi, sqrt

pendulums = [5,6,7,8,9,10,11,12] 
gravity = 10 # m/s**2

lengths = [x**2 for x in pendulums]  # cm

def gcd(*numbers):
    """Return the greatest common divisor of the given integers"""
    from fractions import gcd
    return reduce(gcd, numbers)


def lcm(*numbers):

    """Return lowest common multiple."""    
    def lcm(a, b):
        return (a * b) // gcd(a, b)
    return reduce(lcm, numbers, 1)

T = 2 * pi * sqrt( (lcm(*lengths)/100) / gravity) # Period in seconds using greatest common multiple

print  'Hours until all pendulums are in sync: %0.2f' % ( T/(60*60) ) # convert to hours

Hours until all pendulums are in sync: 1.53

Michael Fitzgerald - 3 years, 6 months ago

Arjen Vreugdenhil - 3 years, 6 months ago

Pendulums, pendulums everywhere

1 solution

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