# To find the length of a pendulum in Simple Harmonic motion!

1. Feb 10, 2010

### Yehia11

If we have 2 simple pendulums, (one longer than the other) oscillating in SHM in step. The next time they are in step is after 20 seconds has elapsed, during which the time the longer pendulum has completed exactly 10 oscillations. Find the string lengths

I found that the length of the long one is 1 metre through the equation:

Periodic Time (which is 2) = 2pi X root(l/g)

but how do we find the length of the longer one??

thankyou in advance. help very appreciated!

2. Feb 11, 2010

### torquil

Is this homework? If so it should be in the homework section.

Torquil

3. Feb 11, 2010

### Yehia11

It is not homework, so no, it should not be in the homework section.

Last edited: Feb 11, 2010
4. Feb 11, 2010

### torquil

Sorry, but I think it is good to ask whenever there is a problem that looks "homework-like". Don't take it personally.

Anyway, the shorter (and faster) pendulum must have completed exactly 11 periods when they are in step for the first time, since the longer one has completed exactly 10.

Thus the pendulum periods satisfy T_long * 10 = T_short * 11.

This gives you T_short which is slightly smaller than T_long, and you can use your formula to calculate its length.

Torquil

5. Feb 11, 2010

### torquil

Just to clarify, you meant to say "shorter one" here, right? Because you have already found the length of the longer one to be approximately 1m. Anyway, if my calculation mixes this up, just modify my numbers a bit (the 10 and 11 in my formula) to fit with the exact problem. The idea is the same.

Torquil

6. Feb 11, 2010

### Yehia11

No problem, sorry if i sounded antagonistic, no harm in asking.

Anyway, firstly YES your right I did mean the SHORTER one, sorry my bad. We know the length of the longer one, it is the shorter one we are trying to find. :)

And secondly why must the shorter one have completed exactly 11 cycles? we know that the long one (and slower one) completed 10, but i feel we have too little info about the short one? or am i wrong? thank you very much for your input, i appreciate it.

7. Feb 11, 2010

### torquil

First, the shorter one must have completed an exact whole number of periods, since the two pendulums meet when the longer one has completed an exact whole number of periods.

In addition, if the shorter one has completed more than one extra period (ie. a total of 12 of more), then they would have been in step much earlier, since it would have had to overtake the longer one at least once before t=20s. So then it would not be true that it was the first time they were in step. Therefore the shorter one must have completed exactly one more period than the longer one.

E.g. if the shorter one had completed 12 in that time, they would have been in step after 10s also, which is not true according to the problem statement. If the shorter one had completed 13, they would have been in step after exactly 20/3 seconds, and so on.

I'm assuming that "in step" means being at the same position with a velocity in the same direction.

Torquil

8. Feb 11, 2010

### Redbelly98

Staff Emeritus