Answer to Question #258306 in Physics for jay

Question

In a simple pendulum experiment to determine g the equation used is

T = 2π √ (l /g)

where T, the period is found to be (2.16 ± 0.01) s when the length l of the pendulum is

(1.150 ± 0.005) m. Find the value of g and its uncertainty.

Answer 1

The value of g:

$$g_t=\frac{4\pi^2 l}{T^2}=9.73\text{ m/s}^2.$$

Maximum uncertainty:

$$\Delta g_\text{max}=\bigg|g_t-\frac{4\pi^2 (1.15+0.005)}{(2.16-0.01)^2}\bigg|=0.13\text{ m/s}^2,\\\space\\ \Delta g_\text{min}=\bigg|g_t-\frac{4\pi^2 (1.15-0.005)}{(2.16+0.01)^2}\bigg|=0.13\text{ m/s}^2,\\\space\\ \Delta g=\frac12(\Delta g_\text{min}+\Delta g_\text{max})=0.13\text{ m/s}^2.\\\space\\ g=g_t\pm\Delta g=(9.73\pm0.13)\text{ m/s}^2.$$