Answer :
The period of motion for the length 0.500 m is 1.42 s. In order to determine the period of motion for each length of the simple pendulum in the given problem, we will use the formula for the period of a simple pendulum, given as;
T = 2π√(l/g)
Where,
T = period of the motion
l = length of the pendulum
g = acceleration due to gravity (9.81 m/s²)
Here, a small object is attached to the end of a string to form a simple pendulum. The period of its harmonic motion is measured for small angular displacements and three lengths, each time clocking the motion with a stopwatch for 50 oscillations.
For lengths of 1.000 m, 0.750 m, and 0.500 m, total times of 99.8 s, 85.5 s, and 71.1 s are measured for 50 oscillations respectively.
So, let's calculate the period of motion for each length of the simple pendulum;(a) For l = 1.000 m,
Total time for 50 oscillations, T = 99.8 s
Period of motion, T = T/50T = 99.8 s/50T = 1.996 s
Now, we will plug the value of l and g into the given formula;
T = 2π√(l/g)
T = 2π√(1.000 m/9.81 m/s²)
T = 2π x 0.3185
T = 2.000 s (approx)
Therefore, the period of motion for the length 1.000 m is 2.000 s.(b) For l = 0.750 m,
Total time for 50 oscillations, T = 85.5 s
Period of motion, T = T/50
T = 85.5 s/50
T = 1.71 s
Now, we will plug the value of l and g into the given formula;
T = 2π√(l/g)T = 2π√(0.750 m/9.81 m/s²)
T = 2π x 0.2741
T = 1.72 s (approx)
Therefore, the period of motion for the length 0.750 m is 1.72 s.
(c) For l = 0.500 m,
Total time for 50 oscillations, T = 71.1 s
Period of motion, T = T/50T = 71.1 s/50T = 1.42 s
Now, we will plug the value of l and g into the given formula;
T = 2π√(l/g)
T = 2π√(0.500 m/9.81 m/s²)
T = 2π x 0.2256
T = 1.42 s
Hence, we have found the main answer to the question and have determined the period of motion for each length of the simple pendulum.
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