The period of simple pendulum on the surface of Mars is about 2.22 s
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Simple Harmonic Motion is a motion where the magnitude of acceleration is directly proportional to the magnitude of the displacement but in the opposite direction.
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The pulled and then released spring is one of the examples of Simple Harmonic Motion. We can use the following formula to find the period of this spring.
[tex]T = 2 \pi\sqrt{\frac{m}{k}}[/tex]
T = Periode of Spring ( second )
m = Load Mass ( kg )
k = Spring Constant ( N / m )
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The pendulum which moves back and forth is also an example of Simple Harmonic Motion. We can use the following formula to find the period of this pendulum.
[tex]T = 2 \pi\sqrt{\frac{L}{g}}[/tex]
T = Periode of Pendulum ( second )
L = Length of Pendulum ( kg )
g = Gravitational Acceleration ( m/s² )
Let us now tackle the problem !
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Given:
Period of the pendulum on the Earth = T_e = 1.35 s
Gravitational Acceleration of the Earth = g_e = 9.8 m/s²
Gravitational Acceleration of the Mars = g_m = 0.37 × 9.8 = 3.626 m/s²
Unknown:
Period of the pendulum on the Mars = T_m = ?
Solution:
[tex]T_e : T_m = 2 \pi\sqrt{\frac{L}{g_e}} : 2 \pi\sqrt{\frac{L}{g_m}}[/tex]
[tex]T_e : T_m = \sqrt{\frac{1}{g_e}} : \sqrt{\frac{1}{g_m}}[/tex]
[tex]T_m = \sqrt{\frac{g_e}{g_m}} \times T_e[/tex]
[tex]T_m = \sqrt{\frac{9.8}{3.626}} \times 1.35[/tex]
[tex]T_m \approx 2.22 \texttt{ s}[/tex]
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Grade: High School
Subject: Physics
Chapter: Simple Harmonic Motion
