Simple pendulum problem - confirmation

[Elysium]

Hi, I got this question:

If a simple pendulum with length 1.50 m makes 72.0 oscillations in 180 s, what is the acceleration of gravity at its location?

Ok, for the period:

$$\frac{180\textit{s}}{72.0} = 2.50{\textit{s}}$$

Then I found the pendulum equation that I don't understand well. I arraganged it to find the gravity:

$$g = \left( \frac{2\pi\sqrt{\ell}}{T} \right) ^2$$

Then I plugged in the numbers and got the answer. Did I do this correctly?

 

[mezarashi]

Looks fine =)

But do feel free to clarify your doubts with the pendulum equation if you have any ^^

 

[quicknote]

Hi there,

Yes, you have solved the simple pendulum problem correctly. The equation you have used is known as the pendulum equation, which relates the period (T) of a pendulum to its length (l) and the acceleration due to gravity (g) at its location. By rearranging the equation and plugging in the given values, you have correctly calculated the acceleration due to gravity at the location of the pendulum.

It is important to note that the value of g may vary slightly depending on the location of the pendulum, as it is affected by factors such as the Earth's rotation and local topography. However, for most practical purposes, a value of 9.8 m/s^2 is commonly used for the acceleration due to gravity.

I hope this helps and keep up the good work in your scientific problem-solving!