Using a small pendulum of length 0.171 m, a geophysicist observes that it completes 1 cycle every 0.833 s. What is the value of g at his location?
T=2πsqrt(L/g),
g=4π²L/T²=4π²•0.171/0.833² =...
To find the value of g at the geophysicist's location, we can use the formula for the period of a pendulum:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
Given that the pendulum completes 1 cycle every 0.833 s and the length is 0.171 m, we can plug these values into the formula and solve for g:
0.833 s = 2π√(0.171 m / g)
Rearranging the equation, we get:
√(0.171 m / g) = 0.833 s / (2π)
Simplifying further:
0.171 m / g = (0.833 s / (2π))^2
Now we can solve for g:
g = 0.171 m / [(0.833 s / (2π))^2]
Calculating the value:
g = 0.0138 m/s²
Therefore, the value of g at the geophysicist's location is approximately 0.0138 m/s².
To find the value of the acceleration due to gravity, g, at the geophysicist's location, we can use the formula for the period of a simple pendulum:
T = 2π√(L / g)
Where:
T = period of the pendulum (in seconds)
L = length of the pendulum (in meters)
g = acceleration due to gravity (in meters per second squared)
In this case, we are given:
L = 0.171 m
T = 0.833 s
We can rearrange the formula to solve for g:
g = (4π²L) / T²
Plugging in the given values:
g = (4π² * 0.171) / (0.833)²
Calculating this expression gives us the value of g at the geophysicist's location.