if a simple pendulum with length 1.50 m makes 72 oscillations in 3 minutes. what is the acceleration of the gravity at its location ?

let I = m L^2

theta = A sin wt
torque = -m g L sin theta = approx -m g L theta
dTheta/dt = A w cos w t
d^2Theta/dt^2 = -Aw^2 sin wt = - w^2 Theta
I d^2Theta/dt ^2 = m g L theta
m L^2 w^2 = m g L
w^2 = g/L
(YOU MIGHT START HERE IF FORMULA IN YOUR BOOK)
2 pi f = sqrt(g/L)
f = 72/180
2 pi (72/180) = sqrt (g/1.5)
6.317 = g/1.5
g = 9.47 m/s^2

Well, let's swing into action and calculate the acceleration of gravity!

The formula to calculate the period (T) of a simple pendulum is T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration of gravity.

In this case, the period T is 3 minutes divided by 72 oscillations which gives us 3/72 minutes per oscillation. But since we need the period in seconds, let's convert that to 3/72 multiplied by 60 seconds per minute.

So, the period T is approximately 2.500 seconds.

Now, let's plug this value into the formula: 2.500 = 2π√(1.50/g).

To simplify, we can divide both sides by 2π: 2.500 / (2π) = √(1.50/g).

And now, let's square both sides: (2.500 / (2π))^2 = 1.50/g.

Simplifying further gives us: (2.500^2 / (2π)^2) = 1.50/g.

Finally, we can rearrange the equation to solve for g: g = 1.50 / (2.500^2 / (2π)^2).

After doing the math, we find the acceleration of gravity at its location to be approximately 9.78 m/s^2.

So, gravity is really pulling its weight in this situation!

To find the acceleration due to gravity at the location of the simple pendulum, you can use the formula for the period of a simple pendulum:

T = 2π√(L/g)

Where:
T = period of the pendulum
L = length of the pendulum
g = acceleration due to gravity

First, let's find the period of the pendulum using the given information.

Given:
Length of the pendulum (L) = 1.50 m
Number of oscillations in 3 minutes = 72

Step 1: Convert 3 minutes to seconds
1 minute = 60 seconds
So, 3 minutes = 3 * 60 = 180 seconds

Step 2: Find the period (T)
T = 180 seconds / 72 oscillations
T = 2.5 seconds

Now, we can use the period to find the acceleration due to gravity (g).

Step 3: Rearrange the formula for the period to solve for g
T = 2π√(L/g)
T^2 = 4π^2(L/g)
g = 4π^2(L/T^2)

Step 4: Substitute the given values into the formula
g = 4 * (3.1416)^2 * (1.50 m) / (2.5 seconds)^2

Step 5: Calculate the value of g
g ≈ 9.78 m/s^2

Therefore, the acceleration due to gravity at the location of the simple pendulum is approximately 9.78 m/s^2.

To find the acceleration due to gravity at the location of the pendulum, we can use the formula for the period of a pendulum, which is given by:

T = 2π√(L/g)

where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

We are given the period in minutes, so we need to convert it to seconds. There are 60 seconds in a minute, so the period in seconds is:

T = 3 minutes * 60 seconds/minute = 180 seconds

We are also given the length of the pendulum, which is 1.50 m.

Now, rearranging the formula for the period, we can solve for g:

g = (4π²L) / T²

Substituting the values we have:

g = (4π² * 1.50) / (180)²

Calculating this expression gives us:

g ≈ 9.811 m/s²

Therefore, the acceleration due to gravity at the location of the pendulum is approximately 9.811 m/s².