In an experiment for determining the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period square is plotted against the string length of the pendulum in the figure. What is the value of g at the place ?

Question

In an experiment for determining the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period square is plotted against the string length of the pendulum in the figure. What is the value of g at the place ? (A) 9.81 m/s2 (B) 9.87 m/s2 (C) 9.91 m/s2 (D) 10.0 m/s2

Tardigrade Admin · Accepted Answer

T =2 π √(ℓ/ g ) T 2=4 π2 (ℓ/ g ) g =4 π2 (ℓ/ T 2) T 2=(4 π2/ g ) (4 π2/ g )=4 g =π2=9.87

Q. In an experiment for determining the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period square is plotted against the string length of the pendulum in the figure.
What is the value of g at the place ?

9.81 m/s²

9.87 m/s²

9.91 m/s²

10.0 m/s²

$T = 2 π \frac{ℓ}{g}$
$T^{2} = 4 π^{2} \frac{ℓ}{g}$
$g = 4 π^{2} \frac{ℓ}{T ^{2}}$
$T^{2} = \frac{4 π ^{2}}{g}$
$\frac{4 π ^{2}}{g} = 4$
$g = π^{2} = 9.87$

Q. In an experiment for determining the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period square is plotted against the string length of the pendulum in the figure. What is the value of g at the place ?

9.81 m/s2

9.87 m/s2

9.91 m/s2

10.0 m/s2