In this example we will use pendulum motion to actually measure the acceleration of gravity on a different planet. An astronaut on the surface of Mars measures the frequency of oscillation of a simple pendulum consisting of a ball on the end of a string. He finds that the pendulum oscillates with a period of 1.5 s. But the acceleration due to gravity on Mars is less than that on earth, gMars=0.38gearth. Later, during a journey to another planet, the astronaut finds that his simple pendulum oscillates with a period of 0.92 s. What planet is he now on?SOLUTIONSET UP Each planet has a different value of the gravitational acceleration g near its surface. The astronaut can measure g at his location, and from this he can determine what planet he's on. First we use the information about Mars to find the length L of the string that the astronaut is swinging. Then we use that length to find the acceleration due to gravity on the unknown planet.

Respuesta :

Answer:

Explanation:

Let length of the pendulum be l . The expression for time period of pendulum is as follows

T = 2Ļ€[tex]\sqrt{\frac{l}{g} }[/tex]

For Mars planet ,

1.5 = [tex]2\pi\sqrt{\frac{l}{.38\times9.8} }[/tex]

For other planet

.92 = [tex]2\pi\sqrt{\frac{l}{g_1} }[/tex]

Squiring and dividing the two equations

[tex]\frac{1.5^2}{.92^2} = \frac{g_1}{3.8\times9.8}[/tex]

[tex]g_1 = 9.9[/tex]

The second planet appears to be earth.

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