Question

At sea level, at a latitude where , a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m. What is the value of g in m/s2 at a location where the length of such a pendulum is 0.970 m

Answer 1

Using the formula for the period of a simple pendulum, the acceleration due to gravity at the new location with a pendulum length of 0.970 m is approximately 9.78 m/s².

Step-by-step explanation:

To calculate the acceleration due to gravity at a different location based on pendulum length and period changes, we can use the formula for the period of a simple 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 period of the pendulum at sea level (where g = 9.8 m/s²) with a length of 0.993 m is 2.00 s, we can solve for g at the new location where the pendulum length is 0.970 m assuming the period remains the same:

g = (4π² × l) / T²

At the new location:

g = (4π² × 0.970) / (2.00)² ≈ 9.78 m/s²

Thus, the acceleration due to gravity at the new location is approximately 9.78 m/s².

Answer 2

a) The value of g at such location is:

`g=9.8005171\,(m)/(s^2)`

b) the period of the pendulum with the length is 0.970 m is:

`T=1.9767 sec`

Step-by-step explanation:

Recall the relationship between the period (T) of a pendulum and its length (L) when it swings under an acceleration of gravity g:

`L=(g)/(4\,\pi^2) \,T^2`

a) Then, given that we know the period (2.0 seconds), and the pendulum's length (L=0.993 m), we can determine g at that location:

`g=(4\,\pi^2\,L)/(T^2)\\g=(4\,\pi^2\,0.993)/((2)^2)\\g=\pi^2\,(0.993)\,(m)/(s^2) \\g=9.8005171\,(m)/(s^2)`

b) for this value of g, when the pendulum is shortened to 0.970 m, the period becomes: