Physics, 11.07.2019 07:30 hollycoleman13

Acertain simple pendulum has a period on the earth of 1.70 s . what is its period on the surface of mars, where g=3.71m/s2?

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Answer from: yhbgvfcd6677

The time period for the simple the simple pendulum on the earth is

$T_{e} =2\pi \sqrt{\frac{l}{g_{e} } }$

The time period for the simple the simple pendulum on the mars is

$T_{m} =2\pi \sqrt{\frac{l}{g_{m} } }$

Here, $T_{e}$ and $T_{m}$ are the time periods on the earth and mars respectively and $g_{e}$ and $g_{m}$ are the acceleration due to gravity on earth and mars respectively.

We can also write,

$\frac{T_{e}}{T_{m}} =\frac{\sqrt{\frac{l}{g_{e}} } }{\sqrt{\frac{l}{g_{m}} } }$

$\frac{T_{e}}{T_{m}} =\sqrt{\frac{g_{m}}{g_{e}} }$

Given $T_{e} = 1.70 s$ and $g_{m} = 3.71 m/s^2$

Therefore,

$\frac{1.70 s}{T_{m} } =\sqrt{\frac{3.71 m/s^2}{9.8 m/s^2} }$

$T_{m}=\frac{1.70}{0.615} = 2.76 s$ .

Thus, the time period on the surface of mars is 2.76 sec.

Answer from: elijahcarson9015

The period of simple pendulum on the surface of Mars is about 2.76 s

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Further explanation

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.

$T = 2 \pi\sqrt{\frac{m}{k}}$

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.

$T = 2 \pi\sqrt{\frac{L}{g}}$

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 = 1.70 s

Gravitational Acceleration of the Earth = g_e = 9.8 m/s²

Gravitational Acceleration of the Mars = g_m = 3.71 m/s²

Unknown:

Period of the pendulum on the Mars = T_m = ?

Solution:

$T_e : T_m = 2 \pi\sqrt{\frac{L}{g_e}} : 2 \pi\sqrt{\frac{L}{g_m}}$

$T_e : T_m = \sqrt{\frac{1}{g_e}} : \sqrt{\frac{1}{g_m}}$

$T_m = \sqrt{\frac{g_e}{g_m}} \times T_e$

$T_m = \sqrt{\frac{9.8}{3.71}} \times 1.70$

$T_m \approx 2.76 \texttt{ s}$

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Learn moreModel for Simple Harmonic Motion : Force of Simple Harmonic Motion : Example of Simple Harmonic Motion :

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Answer details

Grade: High School

Subject: Physics

Chapter: Simple Harmonic Motion

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Keywords: Simple , Harmonic , Motion , Pendulum , Spring , Period , Frequency

Acertain simple pendulum has a period on the earth of 1.70 s . what is its period on the surface of

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