Answer:
18.4 m
Explanation:
The period of a simple pendulum is given by the equation
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex] (1)
where
L is the length of the pendulm
g is the acceleration due to gravity
In this problem, we know that:
N = 13 is the number of oscillations completed by the pendulum
t = 112 s is the time taken to complete those oscillations
So the frequency of the pendulum is:
[tex]f=\frac{N}{t}=\frac{13}{112}=0.116 Hz[/tex]
The period is the reciprocal of the frequency, so for this pendulum:
[tex]T=\frac{1}{f}=\frac{1}{0.116}=8.62 s[/tex]
Now we can use eq(1) to find the length of the pendulum. We know that
[tex]g=9.8 m/s^2[/tex]
Therefore, re-arranging the equation for L, we find:
[tex]L=(\frac{T}{2\pi})^2g=(\frac{8.62}{2\pi})^2 (9.8)=18.4 m[/tex]
The length of the pendulum in the service shaft of a building used for the experiment is 18.4m.
Here we assume the case of a simple pendulum, whose time period of revolution (T) is related to the length of the pendulum (L) as given below:
[tex]T = 2\pi\sqrt{\frac{L}{g} }[/tex]
Now, it is given that the bob is observed to complete 13 oscillations in 112s,
so the frequency of the bob is given by:
f = 13/112
f = 0.116 Hz
The time period T is the inverse of the frequency, so:
T = 1/f
T = 1 / (0.116) s
T = 8.62s
Thus,
[tex]T = 2\pi\sqrt{\frac{L}{g} }\\\\L=\frac{gT}{4\pi^2} \\\\L=\frac{9.8\times8.62}{4\times3.14^2}[/tex]
L = 18.4m
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