At sea level, at a latitude where $g=9.80 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}$ , 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?
A)0.098 3
B)3.05
C)9.57
D)10.0
E)38.3

Question

Quizplus · Accepted Answer

We can start by using the equation 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.

At the first location (sea level, length=0.993 m), we can plug in the values and solve for g:
2 = 2π√(0.993/g)
g = (4π^2)(0.993)/(2^2)
g = 9.86 m/s^2

At the second location (length=0.970 m), we can use the same equation and solve for g:
2 = 2π√(0.970/g)
g = (4π^2)(0.970)/(2^2)
g = 9.57 m/s^2

Therefore, the value of g in m/s^2 at the second location is 9.57, which corresponds to choice C.

At Sea Level, at a Latitude Where g=9.80 m s2g=9.80 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}g=9.80 s2 m​ , a Pendulum That Takes 2

At Sea Level, at a Latitude Where $g=9.80 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}$ , a Pendulum That Takes 2