Physical Pendulum - Theoretical equation

[Physics Curiosity]

Homework Statement

Question requires you to 'process' the independent variable, (the thing that's manipulated thing on a pendulum),

Relevant Equations

The simple pendulum equation is T = 2π√(L/g), in which the independent variable "L" is 'processed' into T = 2π/√(g) * √(L). I am wondering how in this equation, T = 2π√(2L/3g), how the 2L can be processed into √(L) like the first example of the simple pendulum equation being processed?

Initially I went from:

T = 2π√(2L/3g)
T = 2π/√(3g) * √(2L)
To finally this equation:
T = 2π/√(3g) * √(L)
Where 2L becomes L as the 2 is lost. I am not fully sure if this is correct or how to properly get rid of the 2 in 2L.We must follow the rule of y = mx+c whereby y = T, m = the constant variables (pi and g), and x is the independent variable (L). There is no y-intercept c. Hence the equation must be 'processed' in this way.

 

[BvU]

Hello curious,

$\qquad$ !​

First things first :

The equation $$T=2\pi\sqrt{L\over g}$$ is for a mathematical pendulum and is valid for small amplitudes.

An experimental determination of the acceleration from gravity can be to measure a single $T$ for a single value of $L$.
An improvement is then to vary $L$ and measure a series of $T$ for these. The expected dependency is then $$T={2\pi\over {\sqrt g}}\; \sqrt{L} $$ and a plot of $T$ as a function of $\sqrt L$ is expected to yield a straight line through the origin with a slope of ${2\pi\over {\sqrt g}}$.

Does that answer your question ?

$\$

 

[Physics Curiosity]

Thanks for the further clarification!

I do know how to manipulate that equation in that way, except I am confused on how I can process with T = 2πsqrt(2L/3g), I am not sure how to get rid of the 2 in 2L. I came to this T = 2π/√(3g) * √(2L), but the √(2L) should be √(L) instead.

 

[haruspex]

I am not at all sure what is meant by "processed" here.
At a guess, what you mean is turning an expression which has an L inside a square root into the form $(expression)\sqrt L$. If so, doesn't look hard; just leave the 2 where it was.
In case I misunderstand, please post the question exactly as given to you.

 

[Physics Curiosity]

I am supposed to be graphing sqrt(L), not sqrt(2L). Here's an example of "m" being "processed", I am guessing processed just means to separate the equation into y = mx+c structure, where T is period, m is the gradient, x is the independent variable (m), and c is the y-intercept. This example however may not be applicable to the T = 2πsqrt(2L/3g) equation I gave earlier, if so please confirm if possible, thanks! I am assuming T = 2π/√(3g) * √(2L) is the correct way?

 

[haruspex]

I am supposed to be graphing sqrt(L), not sqrt(2L). Here's an example of "m" being "processed", I am guessing processed just means to separate the equation into y = mx+c structure, where T is period, m is the gradient, x is the independent variable (m), and c is the y-intercept.View attachment 327727

You mean "x is the independent variable (√m)", right?
Ok, that accords with my guess in post #4, so as I wrote, just leave the 2 where it was.

 

[Physics Curiosity]

I understand now.

I had confused myself -- T = 2π/√(3g) * √(2L) was the correct "processed" equation.

Cheers, many thanks for all the help guys!

 

[DrClaude]

I am supposed to be graphing sqrt(L), not sqrt(2L). Here's an example of "m" being "processed", I am guessing processed just means to separate the equation into y = mx+c structure, where T is period, m is the gradient, x is the independent variable (m), and c is the y-intercept. This example however may not be applicable to the T = 2πsqrt(2L/3g) equation I gave earlier, if so please confirm if possible, thanks! I am assuming T = 2π/√(3g) * √(2L) is the correct way?
View attachment 327727

My reading of this is that "processing" consists here in taking the square root of the value, not isolating it in the original equation. Imagine you have, in some plotting software, a column with values of T and another with values of m. Before you can make the plot (if you want the result to be linear), you need to process $m$ to $\sqrt{m}$, i.e., take the square root of the values of m.

T = 2π√(2L/3g)
T = 2π/√(3g) * √(2L)
To finally this equation:
T = 2π/√(3g) * √(L)
Where 2L becomes L as the 2 is lost. I am not fully sure if this is correct or how to properly get rid of the 2 in 2L.

You cannot remove the 2 like that. Your mistake, on order to get what you want, is in isolating $\sqrt{2L}$ and not simply $\sqrt{L}$. Remember that the square root distributes over multiplications and divisions, hence
$$
\sqrt{\frac{2L}{3g}} = \frac{\sqrt{2}\sqrt{L}}{\sqrt{3}\sqrt{g}},
$$
or more simply
$$
\sqrt{\frac{2L}{3g}} = \sqrt{\frac{2}{3g}} \sqrt{L}.
$$

 

[Physics Curiosity]

In that case would the theoretical gradient, or m, be √(2/3g), then x would be sqrt(L), in the y=mx+c equation?

 

[DrClaude]

In that case would the theoretical gradient, or m, be √(2/3g), then x would be sqrt(L), in the y=mx+c equation?

Yes, with the exception that there is a $2\pi$ missing in the slope m.

 

[Physics Curiosity]

So, I cant just use T = 2π/√(3g) * √(2L), the theoretical gradient being 2π/√(3g) which follows the y=mx+c format like post #6 suggested?

I will be comparing the theoretical gradient to the experimental gradient when I graph √(L) in √(m) in the x-axis against Period T in (s) in the y-axis.

 

[DrClaude]

So, I cant just use T = 2π/√(3g) * √(2L), the theoretical gradient being 2π/√(3g) which follows the y=mx+c format like post #6 suggested?

Only if you plot $T$ as a function of $\sqrt{2L}$. But if what you have is $T$ as a function of $\sqrt{L}$, then the slope is not 2π/√(3g).

 

[Physics Curiosity]

Alright, I am still rather puzzled where the 2pi went, but thanks a bunch for the information. I will look into it!

 

[haruspex]

So, I cant just use T = 2π/√(3g) * √(2L), the theoretical gradient being 2π/√(3g) which follows the y=mx+c format like post #6 suggested?

That is not what I suggested.
You have $$T=2\pi\sqrt{\frac{2L}{3g}}$$
I said to move the $\sqrt L$ out, leaving the 2 where it was. That produces $$T=2\pi\sqrt{\frac{2}{3g}}\sqrt L$$
If you then plot $T$ against $\sqrt L$ you should get a gradient $2\pi\sqrt{\frac{2}{3g}}$.
To get a gradient $\frac{2\pi}{\sqrt{3g}}$ you would need to plot $T$ against $\sqrt {2L}$

 

[Steve4Physics]

Hi @Physics Curiosity. Can I add that it is common (in my experience) to ‘square’ these sorts of equations rather than use square roots. For example...

(To avoid ambiguity, upper case $M$ = mass and lower case $m$ = gradient of graph.)

For a mass oscillating on a spring: $T = 2\pi \sqrt {\frac Mk}$

Squaring both sides: $T^2 = 4\pi^2 \frac Mk$

Rearrange as: $T^2 = \frac {4\pi^2}k ~M$

For a graph of $T^2$ (on the 'y'-axis) against $M$ (on the 'x'-axis) the gradient ($m$) is $m = \frac {4\pi^2}k$.

For example if you wanted to determine $k$ you would measure the gradient then: $k = \frac {4\pi^2}m$

But if you have been told to do it with square roots, that's what you should do.

 

[Physics Curiosity]

That is not what I suggested.
You have $$T=2\pi\sqrt{\frac{2L}{3g}}$$
I said to move the $\sqrt L$ out, leaving the 2 where it was. That produces $$T=2\pi\sqrt{\frac{2}{3g}}\sqrt L$$
If you then plot $T$ against $\sqrt L$ you should get a gradient $2\pi\sqrt{\frac{2}{3g}}$.
To get a gradient $\frac{2\pi}{\sqrt{3g}}$ you would need to plot $T$ against $\sqrt {2L}$

Apologies for my misunderstanding. It was that simple, I just had a blind eye to the question. You have fully answered my question.

I sincerely thank anyone who has taken time out of their day to guide my misunderstanding.

Have a great day! :-)