# Homework Help: Number of oscillations executed by an object to be in phase

1. Jun 22, 2014

### desmond iking

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. 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 dont understand it. can someone explain please?

2. Jun 22, 2014

### CWatters

Try drawing it.

3. Jun 22, 2014

### desmond iking

here's the image

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4. Jun 22, 2014

### CWatters

No I meant draw two appropriate sine waves on the same axis.

5. Jun 22, 2014

### rude man

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.

6. Jun 23, 2014

### desmond iking

what do you mean by n2=n1 +1 ? I can't understand

7. Jun 23, 2014

### rude man

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.

8. Jun 23, 2014

### Staff: Mentor

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 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

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: Jun 23, 2014
9. Jun 24, 2014

### desmond iking

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

how can this relate to the question?

10. Jun 24, 2014

### Staff: Mentor

Therefore, the values of x and y are ...?

11. Jun 24, 2014

### desmond iking

52200

12. Jun 24, 2014

### desmond iking

But the 52200 doesnt comply with the ans

13. Jun 24, 2014

### Staff: Mentor

14. Jun 24, 2014

### desmond iking

x=145, y=144

15. Jun 24, 2014

### Staff: Mentor

http://imageshack.com/a/img29/6853/xn4n.gif [Broken]

Last edited by a moderator: May 6, 2017
16. Jun 24, 2014

### desmond iking

Then wat does it mean? O:)