[SPLIT] The period of a pendulum varies directly as the

[georgebaseball]

Please check my work on this exercise:

The period (the time required for one complete oscillation) of a simple pendulum varies directly as the square root of its length. If a pendulum 12 feet long has a period of 4 seconds, find the period of a pendulum of length 3 feet.

The equation would be:

. . .t = √L / p

So:

. . .√12 / 4 = 0.08

To find the period of a pendelum of lenght 3 feet:

. . .√3 * 0.8 = 1.3 seconds

Is that right?
thanks

 

[stapel]

What do your variables stand for? Where is your variation constant "k"?

Thank you.

Eliz.

 

[georgebaseball]

Good morning stapel
thanks for taking the time of reading and help me to solve my h.w



I realized I had done something wrong about that exercise, i had divided when I had to multiply, so this is how it looks after I corrected it, please tell me if i'm wrong

t= √L * K(s)

t stands por period , √L stands for square root of the lenght, k is the constand and s stands for seconds
then

t= (4) * k √ 12
4 = 3.464 k
k = 1.154

t = √ 3 * 1.154 = 1.99 seconds

is that right????

 

[skeeter]

your final solution is correct, but put away that calculator ...

\(\displaystyle \L T = k\sqrt{L}\)

\(\displaystyle \L 4 = k\sqrt{12}\)

\(\displaystyle \L k = \frac{4}{\sqrt{12}} = \frac{2}{\sqrt{3}}\)

\(\displaystyle \L T = \frac{2}{\sqrt{3}}\sqrt{L}\)

\(\displaystyle \L T = 2\sqrt{\frac{L}{3}}\)

now ... substitute 3 for L and find T. Exact solutions are so much nicer, don't you agree?

 

[georgebaseball]

you're right skeeter
I really appreciate your help.