Find the period of a simple pendulum, if
this experiment is performed inside a satellite?
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In a freely orbiting satellite which is revolving around the earth, the net value of acceleration due to gravity is zero and hence we experience zero gravity inside the spacecraft.
And, the time period of a simple pendulum is is inversely proportional to the square root of acceleration due to gravity. Time period=2 ×pi× (l/g)^1/2 (where l is length of the simple pendulum and g is acceleration due to gravity)
The value of g is zero inside a spacecraft, and in the equation g is in the denominator. And any number divided by zero is NOT DEFINED.
As per me, either the time period is not defined or infinity.
And since the time period cannot be not defined(think of it yourself), the answer is infinity which also implies that the pendulum won't oscillate.
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And, the time period of a simple pendulum is is inversely proportional to the square root of acceleration due to gravity. Time period=2 ×pi× (l/g)^1/2 (where l is length of the simple pendulum and g is acceleration due to gravity)
The value of g is zero inside a spacecraft, and in the equation g is in the denominator. And any number divided by zero is NOT DEFINED.
As per me, either the time period is not defined or infinity.
And since the time period cannot be not defined(think of it yourself), the answer is infinity which also implies that the pendulum won't oscillate.
Suggestions are welcome!
If you like please mark brainliest
yasir20:
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are they need a valu for this question or something
I didn't hear not defined
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