pendulum length period

[SaintJude]

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if length of arm holding a pendulum is decreased in half, what happens the period? Does the period decrease by a factor of 0.7 or 1.4 ?

 

[chiddler]

if length of arm holding a pendulum is decreased in half, what happens the period? Does the period decrease by a factor of 0.7 or 1.4 ?

T = 2(pi) sqrt(L/g)

if it helps, i remember this formula via "2 pi's are large", (L/G). teehee!

so it's halved, and square root of half. 1/1.4 = 0.7

 

[SaintJude]

Yeah, Kaplan said decrease by a factor of 1.4, so that must have been a mistake then ?

 

[dsoz]

0.7 = 1/1.4

0.7 and 1.4 are reciprocals of each other...

When something decreases by a factor of 0.7 is actually increasing... (if you divide by 0.7 you get a larger number!)

So Kaplan states it DECREASES by a factor of 1.4, it is also CHANGING by a factor of 0.7

It is all in the semantics.

dsoz

 

[pfaction]

If I may offer my opinion:

T = 2pi rad (Length/Gravity)

Therefore:

T is proportional to rad L.
if: L increased by 2, T increases by rad2 = 1.4
If: L decr by 2,
T decr by rad 2
T decr by 1.4.

 

[SaintJude]

dsoz. Thanks. But how did you know that?

 

[dsoz]

dsoz. Thanks. But how did you know that?

Sometimes I see patterns in numbers, and experience has shown me that 0.7 and 1.4 are approximate reciprocals. (0.7 = 1/1.428... from a calculator).

The key word in the origional problem is DECREASE by a factor. If you divide a number by a decimal you get a larger number, not a decrease. If you are looking for a decrease, then you take the reciprocal of the decimal (1 divided by 0.7) and it gives 1.4 as an answer. That is the facor of the decrease.

If the question was just how much it CHANGED, then 0.7 would be correct.

Like I said, it is semantics.

dsoz

 

[pfaction]

Was my answer wrong?

 

[MedPR]

Yeah, Kaplan said decrease by a factor of 1.4, so that must have been a mistake then ?

No this is correct. When you do these problems you should be canceling out the constants. 2pi is a constant, so it doesn't matter because it has the same value before and after you change something.

Say L was 40. Period would be 4pi.

Now L is 20. Period is 2pi*1.4 = 2.8pi

4/2.8 is 1.4.

If you don't want to use numbers, you can use two equations to see that everything but L cancels out (in this case).

T=2pi(sqrtL/g)
T*=2pi(sqrtL*/g)

L*=L/2

So T*=2pi(sqrtL/2g)
T*=(1/1.4)(2pi)(sqrtL/g)

T*=(1/1.4)(T)

1.4T*=T

T=1.4T*

So T is 1.4 times greater than T*. In other words, T decreased by 1.4.

 

[chiddler]

yes i'm glad dsoz explained my mistake.

very subtle.

 

[SaintJude]

Wow, I can't believe I never noticed this. That was really helpful. But it becomes clear to me when I use simpler number

T* = 1/3T
T* = (T/3)
Here the period has changed by a factor of 1/3. And this is the same as saying that the period has also decreased by a factor of 3.

Similarly

T* = 0.7 T
T* = 1/1.4T
T* = T/ 1.4 The period has changed by a factor of 0.7 and decreased by a factor of 1.4...

Wow, thanks again.

 

[MedPR]

Wow, I can't believe I never noticed this. That was really helpful. But it becomes clear to me when I use simpler number

T* = 1/3T
T* = (T/3)
Here the period has changed by a factor of 1/3. And this is the same as saying that the period has also decreased by a factor of 3.

Similarly

T* = 0.7 T
T* = 1/1.4T
T* = T/ 1.4 The period has changed by a factor of 0.7 and decreased by a factor of 1.4...

Wow, thanks again.

That's true but you might be missing the point of this problem. sqrt2 is 1.4 and they were testing if you would divide by the 2 in 2pi or if you would recognize that it falls off so the answer is just 1.4. The fact that 0.7 is about equal to 1/1.4 is a coincidence in this case.

They were testing if you knew that the answer was sqrt2 and not (sqrt2)/2.