Answer:
5.63m/s2
Step-by-step explanation:
Step 1:
Data obtained from the question.
Length (L) = 1.20 m
Time (t) =290 seconds
Number of oscillation (n) = 100
Period (T) =?
Acceleration due to gravity (g) =?
Step 2:
Determination of the period. This is illustrated below:
Period is the time taken to make one complete oscillation. It is given by:
Period (T) = time (t) / number of oscillation (n)
T = t/n
t = 290 secs
n = 100
T = t/n
T = 290/100
T = 2.9 secs
Step 3:
Determination of the acceleration due to gravity of the planet. This is illustrated below:
Applying the equation T = 2π√(L/g), the value of 'g' can be obtained as follow :
T = 2π√(L/g)
T = 2.9 secs
L = 1.20 m
g =?
T = 2π√(L/g)
2.9 = 2π√( 1.2/g)
Take the square of both side
8.41 = 4π^2 (1.2/g)
8.41 = 4.8π^2 /g
Cross multiply to express in linear form
8.41 x g = 4.8π^2
Divide both side by 8.41
g = 4.8π^2/8.41
g = 5.63m/s2
Therefore, the acceleration due to gravity of the planet is 5.63m/s2