# Math Help - Period of oscillation T = 2π √ l/g?

1. ## Period of oscillation T = 2π √ l/g?

This is a question on a diploma i'm about to complete, i've been fine with everything else, but this one really has me stumped. Any help would be appreciated.

The forumla which relates the period of oscillation, T seconds, of a pendulum of length l metres is T = 2π √ l/g where g is acceleration due to gravity.

a) Find the period of oscillation of a pendulum 1.25m long (Take the value of g to be 9.81 ms^-2

b) Rearrange the formula to make l the subject.

2. Originally Posted by Elrithral
This is a question on a diploma i'm about to complete, i've been fine with everything else, but this one really has me stumped. Any help would be appreciated.

The forumla which relates the period of oscillation, T seconds, of a pendulum of length l metres is T = 2π √ l/g where g is acceleration due to gravity.

a) Find the period of oscillation of a pendulum 1.25m long (Take the value of g to be 9.81 ms^-2
Do the arithmetic! Since g= 9.81 and l= 1.25, that is $2\pi\sqrt{\frac{1.25}{9.81}}$

b) Rearrange the formula to make l the subject.
"Unpeel" the l- that is, do the opposite of what ever has been done to l.
Start from $T= 2\pi\sqrt{\frac{l}{g}}$.

Divide both sides by $2\pi$: $\frac{T}{2\pi}= \sqrt{\frac{l}{g}}$.

Square both sides: $\frac{T^2}{4\pi^2}= \frac{l}{g}$.

Finally, multiply both sides by g: $\frac{T^2 g}{4\pi^2}= l$.

3. Originally Posted by HallsofIvy
Do the arithmetic! Since g= 9.81 and l= 1.25, that is $2\pi\sqrt{\frac{1.25}{9.81}}$
Thanks for the reply. I'd got that far, but my calculations seem stupid.

T = 2π x 0.36
T = 9.87 x 0.36
T = 3.55

That's how I worked it out and it just doesn't seem right. Sorry.

4. Originally Posted by Elrithral
Thanks for the reply. I'd got that far, but my calculations seem stupid.

T = x 0.36
T = 9.87 x 0.36
T = 3.55

That's how I worked it out and it just doesn't seem right. Sorry.
HI

2 pi = 9.87 ?? Check with ur calculator . That should give you 6.28 so

T=6.28 x 0.36