# Thread: I need help with this question.

1. ## I need help with this question.

the question is:

The time, T, in seconds, that a pendulum takes to do a complete oscillation is given by the
formula
T = 2PI
where l is the length of the pendulum, in metres, and g is the acceleration due to gravity.
Take the value of g to be 9.807 m/s2.
In St. Isaac’s cathedral in St. Petersburg there is a pendulum of length 94 m.
18 (a) (i) Calculate the value of T for this pendulum.
Give all the figures in your calculator display.

Can somebody explain to me how to do this,i already know the answer but i want to know how to do it.(This is a past GCSE)Thanks.

2. Are you sure you haven't posted your formula wrong ?
Because the real formula is :

$T = 2 \pi \sqrt{\frac{L}{g}}$ where $L$ is the length of the pendulum and $g \approx 9.807$.

Anyway, you are given the length of the pendulum, and you know the constant $g$. Just plug the values into your equation to find $T$ for this pendulum.

So, you find $T = 2 \pi \sqrt{\frac{94}{9.807}} \approx 19,452512699185704931509913432719$ seconds

Yes I have a lot of figures because I use a high-precision calculator on your calculator you should have the answer rounded around 10 decimal digits. Thus, you find $T \approx 19,45251269$ seconds.

I do not get the last question. $\pi$ is irrationnal, so you cannot express $T$ as a fractional number. Do they just mean round it off sensibly ?

3. Originally Posted by Bacterius
Are you sure you haven't posted your formula wrong ?
Because the real formula is :

$T = 2 \pi \sqrt{\frac{L}{g}}$ where $L$ is the length of the pendulum and $g \approx 9.807$.

Anyway, you are given the length of the pendulum, and you know the constant $g$. Just plug the values into your equation to find $T$ for this pendulum.

So, you find $T = 2 \pi \sqrt{\frac{94}{9.807}} \approx 60,224270304362305376871311925823$ seconds

Yes I have a lot of figures because I use a high-precision calculator on your calculator you should have the answer rounded around 10 decimal digits. Thus, you find $T \approx 60,22427030$ seconds.

I do not get the last question. $\pi$ is irrationnal, so you cannot express $T$ as a fractional number. Do they just mean round it off sensibly ?
Yeh i just copy and pasted that and never realy read over it lol but yeh it did mean
$T = 2 \pi \sqrt{\frac{L}{g}}$i couldnt be botherd to get a calculator xD so tryed to do it in my head.

it sais calc the value of T give all the figs in calcs display and to give your answer as a decimal.

Let me go get my calculator to make my life a bit easyer

4. You should get a calculator, it is useful ...
Otherwise, I answered the part on the calculator's display (just to it on your calculator and copy on your answer booklet), but I don't know what you mean by giving the answer as a decimal. No additional information ?

5. I just looked at the answer they give on the site and its 19.4525127 which is what i got when i just did it in my calculator. so 60.(loads of numbers) is rong?

6. Oh yeah I forgot to do the square root in my calculations, my mistake , that answer is correct
I'll edit my previous posts, thanks for noticing.

7. Could you tell me how i calculate the length of a pendulum that will give a value of T=1?

8. You need to solve this equation for $L$ :

$1 = 2 \pi \sqrt{\frac{L}{g}}$

Extract annoying numbers :

$\frac{1}{2 \pi} = \sqrt{\frac{L}{g}}$

Take the square of both sides to get rid of the square root :

$(\frac{1}{2 \pi})^2 = \sqrt{\frac{L}{g}}^2$

$\frac{1}{4 \pi^2} = \frac{L}{g}$

Multiply by $g$ on both sides :

$\frac{1}{4 \pi^2} \times g = \frac{L}{g} \times g$

$\frac{g}{4 \pi^2} = L$

So the length of a pendulum with a period of $1$ second would be $L = \frac{g}{4 \pi^2} \approx 0.248414$ metres, so about $25$ cm

9. Originally Posted by Ben1191
Could you tell me how i calculate the length of a pendulum that will give a value of T=1?
$T=2\pi \sqrt{\frac{L}{g}}$
square both sides
$T^2=4\pi^2\frac{L}{g}.$
therefore,
$L=\frac{T^2\times g}{4\pi ^2}$

10. A bit late Raoh

11. Originally Posted by Bacterius
A bit late Raoh
I'm too slow

12. Yet you made me realize that I didn't have to isolate the square root to remove it since there are only multiplications !
Looking for an excuse ...
Let's say I did it for the example