Answer:
12.4 m/s²
Explanation:
L = length of the simple pendulum = 53 cm = 0.53 m
n = Number of full swing cycles = 99.0
t = Total time taken = 128 s
T = Time period of the pendulum
g = magnitude of gravitational acceleration on the planet
Time period of the pendulum is given as
[tex]T = \frac{t}{n}[/tex]
[tex]T = \frac{128}{99}[/tex]
T = 1.3 sec
Time period of the pendulum is also given as
[tex]T = 2\pi \sqrt{\frac{L}{g}}[/tex]
[tex]1.3 = 2(3.14) \sqrt{\frac{0.53}{g}}[/tex]
g = 12.4 m/s²
The magnitude of the gravitational acceleration (g) on this planet is equal to 12.52 [tex]m/s^2[/tex].
Given the following data:
Length = 53.0 cm to m = 0.53 m.
Number of cycle, n = 99.0 cycles.
Time = 128 seconds.
First of all, we would calculate the period for a full swing cycle as follows:
[tex]T=\frac{time}{n} \\\\T=\frac{128}{99.0}[/tex]
Period, T = 1.293 seconds.
Mathematically, the time taken (period) by a pendulum is given by this formula:
[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex]
Making g the subject of formula, we have:
[tex]g=\frac{4\pi^2L}{T^2} \\\\g=\frac{4(3.142)^2 \times 0.53}{1.293^2} \\\\g=\frac{20.929}{1.672}[/tex]
g = 12.52 [tex]m/s^2[/tex].
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