Number of oscillations executed by an object to be in phase

[desmond iking]

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?

 

[CWatters]

Try drawing it.

 

[desmond iking]

here's the image

 

[CWatters]

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

 

[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.

 

[desmond iking]

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

 

[rude man]

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.

 

[NascentOxygen]

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 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.

 

[desmond iking]

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.

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

how can this relate to the question?

 

[NascentOxygen]

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 ...?

 

[desmond iking]

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

52200

 

[desmond iking]

But the 52200 doesn't comply with the ans

 

[NascentOxygen]

52200

Re-read https://www.physicsforums.com/showpost.php?p=4781793&postcount=8

 

[desmond iking]

But the 52200 doesn't comply with the ans

x=145, y=144

 

[NascentOxygen]

x=145, y=144

http://imageshack.com/a/img29/6853/xn4n.gif

 

[desmond iking]

Then wat does it mean? O:)