What is the free-fall acceleration in a location
where the period of a 0.594 m long pendulum
is 1.55 s s?
Answer in units of m/s^2
Show equation too please.
period=2PIsqrt(length/g)
square both sides...
T^2=(2PI)^2 l/g
g= (2PI)^2 Length/period^2
I kinda have no clue what you responded.
The equation I have is:
T=2(Pie)*Sqt(L/G)
I need to re-arrange that..
I would square it first, as I did, then rearrange.
By following your equation I got:
5.797880106
If I rearrange it would be:
T^2 = 2(Pie)L/G
then what?
You did not square 2PI.
T^2=(2pi)^2 l/g
g= (2PI)^2 l/t^2
Last question.. sorry
PL = Pie?
I keep getting the question wrong. I think I put it wrong in my caculator.
I put
(2<Pie>.594)^2.594/(1.55^2)
2PI= 6.28
(2PI)^2=about 39, check it
.594=l
1.55^2= about 2.4 check it.
g=39*.594/2.4 in my head, about 9.7 check it.
To find the free-fall acceleration in a specific location using a pendulum, you can use the equation for the period of a pendulum:
T = 2π√(L/g)
Where:
T is the period of the pendulum (in seconds),
L is the length of the pendulum (in meters), and
g is the free-fall acceleration (in m/s^2).
In this case, you are given the period of the pendulum (T = 1.55 s) and the length of the pendulum (L = 0.594 m). You need to solve the equation for g.
Rearranging the equation, we get:
g = (4π²L) / T²
Now, substitute the given values into the equation:
g = (4π² * 0.594) / (1.55)²
Calculating this expression will give you the value of g, the free-fall acceleration in m/s^2.