Number of oscillations executed by an object to be in phase

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Homework Help Overview

The problem involves a turntable rotating at 36.25 rpm and a simple pendulum performing simple harmonic motion (SHM) at 36.00 rpm. The objective is to determine how many complete oscillations the pendulum would execute for both to be in phase.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the periods of the turntable and the pendulum, questioning how to find the point at which they are in phase again. Some suggest using the lowest common multiple (LCM) of their frequencies, while others explore integer relationships between their oscillations.

Discussion Status

The discussion is ongoing, with participants exploring different mathematical approaches to the problem. Some have provided guidance on using LCM and integer relationships, while others are seeking clarification on specific terms and concepts.

Contextual Notes

There is some confusion regarding the definitions and relationships between the variables involved, particularly concerning the integer values representing the number of rotations and oscillations. Participants are also working with the assumption that the pendulum and turntable start in phase.

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Homework Statement



the turntable is rotating at 36.25 rpm. the simple pendulum is performing SHM at 36.00 prm. how many complete oscillations would the pendulum execute so that both are in phase?

Homework Equations





The Attempt at a Solution


my working is 36prm divided by 0.25 rpm =144 oscillations. i just 'simply' work out the answer. and the answer is same with the sample answer coincidently. but i don't understand it. can someone explain please?
 
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Try drawing it.
 
here's the image
 

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No I meant draw two appropriate sine waves on the same axis.
 
The assumption is that they're in phase to begin with, the question is how long before they're in phase again.

You obviously need an integer number of periods of each oscillation.

Turntable's period is T1. Pendulum's period is T2. T2 > T1 here.

Let n1 = number of rotations of the turntable and n2 = no. of pendulum swings. So n1 and n2 are integers

So n1*T1 = n2*T2
but also n1 = n2 + 1 (we want the first time the two are in sync again).
So solve for n2.

You can substitute f1 = 1/T1 and f2 = 1/T2 where f1 = 36.25 rpm and f2 = 36 rpm.
 
what do you mean by n2=n1 +1 ? I can't understand
 
desmond iking said:
what do you mean by n2=n1 +1 ? I can't understand

I defined that. And I said n1 = n2 + 1, not what you quoted.

Think of two sets of domino tiles. One set of dominos has length T1, the other has length T2, where T1 = 1/f1 and T2 = 1/f2 and the T2 tiles are larger than the T1 tiles.

Now start laying the T1 tiles in a row adjacent to a second row of T2 tiles. The two rows are lined up at the start.

When the rows line up again there are 1 more T1 tiles than T2 tiles. So n1 = n2 + 1. n1 is the number of T1 tiles and n2 is the number of T2 tiles.
 
desmond iking said:
the turntable is rotating at 36.25 rpm. the simple pendulum is performing SHM at 36.00 prm. how many complete oscillations would the pendulum execute so that both are in phase?
Work out the lowest common multiple of 36.25 and 36, and from that you will get your answer.

Okay, okay, to appease the mathematicians :smile: I'll rephrase that. First, find the LCM of 3625 and 3600. (Then you'll divide that answer by 100.)

Mathematically, you can write it this way ...

There exists an integer M such that it is simultaneously a multiple of 3600 and of 3625, i.e.,

M = 3600*x = 3625*y[/color]

Find the smallest integers x and y that make the equality in blue true, and integer.

Can you do that? Show your working.
 
Last edited:
nascentoxygen said:
work out the lowest common multiple of 36.25 and 36, and from that you will get your answer.

Okay, okay, to appease the mathematicians :smile: I'll rephrase that. First, find the lcm of 3625 and 3600. (then you'll divide that answer by 100.)

mathematically, you can write it this way ...

There exists an integer m such that it is simultaneously a multiple of 3600 and of 3625, i.e.,

m = 3600*x = 3625*y[/color]

find the smallest integers x and y that make the equality in blue true, and integer.

Can you do that? Show your working.

3600- 3600,7200,10800, 522000
3625- 3625, 7250, 10875, 522000

how can this relate to the question?
 
  • #10
desmond iking said:
3600- 3600,7200,10800, 522000
3625- 3625, 7250, 10875, 522000

how can this relate to the question?
Therefore, the values of x and y are ...?
 
  • #11
nascentoxygen said:
therefore, the values of x and y are ...?

52200
 
  • #12
But the 52200 doesn't comply with the ans
 
  • #14
desmond iking said:
But the 52200 doesn't comply with the ans

x=145, y=144
 
  • #15
desmond iking said:
x=145, y=144

http://imageshack.com/a/img29/6853/xn4n.gif
 
Last edited by a moderator:
  • #16
Then wat does it mean? O:)
 

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