Respuesta :
Answer: 22.8 m
Explanation:
The x-coordinate of the shadow as a function of time t is given by the expression
[tex]x(t)=57 cos(\pi t)[/tex] (1)
where
t is in seconds
x is in centimeters
Here we want to know the distance travelled by the shadown in 20 seconds.
First of all, we find the period of this oscillation. The periodic function can be rewritten as
[tex]x(t)=Acos(\omega t)[/tex] (2)
where
A = 57 cm is the amplitude
[tex]\omega=\frac{2\pi}{T}[/tex] is the angular frequency, with T being the period
By comparing (1) with (2), we notice that
[tex]\frac{2\pi}{T}=\pi\\T=2s[/tex]
So, the period is 2 seconds.
We also know that the amplitude of the motion is 57 cm: this means that in a quarter of period, the shadow covers a distance of 57 cm. So in one period, the distance covered by the shadow is 4 times the amplitude:
[tex]d(1T)=4A=4(57)=228 cm[/tex]
Here we want to find the distance covered by the shadow in 20 seconds, which corresponds to
[tex]n=\frac{t}{T}=\frac{20 sec}{2 sec}=10[/tex]
So, 10 periods. Therefore, the distance covered in 10 periods is:
[tex]D=10(228 cm)=2280 cm = 22.8 m[/tex]
The total distance covered by the shadow of the pendulum in 20 seconds is 22.8 m.
Pendulum and Simple harmonic motion:
The instantaneous amplitude of the shadow of the pendulum is given as:
x(t) = 57cos(πt)
comparing it to the standard equation of Simple harmonic motion:
x(t) = Acos(ωt)
where A is the maximum amplitude
and, ω is the angular frequency of oscillation
we get:
A = 57cm
ω = π rad/s
The time period of oscillation is given by:
T = 2π/ω
T = 2π/π
T = 2s
In one time period, the pendulum completes one full oscillation, the total distance covered by the shadow of the pendulum = 4A = 4×57cm = 228cm = 2.28m
The pendulum takes 2 seconds to complete one oscillation, so in 20 seconds the number of oscillations is:
n = 20/2 = 10
So, the total distance covered = 10 × 2.28m = 22.8m
Learn more about Simple harmonic motion:
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