Number of oscillations executed by an object to be in phase

  • Thread starter Thread starter desmond iking
  • Start date Start date
  • Tags Tags
    Oscillations Phase
AI Thread Summary
The discussion revolves around determining how many complete oscillations a simple pendulum must execute to be in phase with a turntable rotating at 36.25 rpm while the pendulum performs simple harmonic motion (SHM) at 36.00 rpm. The solution involves finding the least common multiple (LCM) of the two frequencies, which helps identify when both systems will align again. Participants clarify the relationship between the number of rotations of the turntable and the pendulum swings, establishing that the turntable completes one more rotation than the number of pendulum swings when they are in sync. The final answer requires calculating specific integer values that satisfy the frequency relationship. Understanding this concept is essential for solving similar phase-related problems in physics.
desmond iking
Messages
284
Reaction score
2

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?
 
Physics news on Phys.org
Try drawing it.
 
here's the image
 

Attachments

  • DSC_0008.JPG
    DSC_0008.JPG
    15.7 KB · Views: 385
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:)
 
Back
Top