For small amplitudes, the period can be calculated as 2 x pi x square root of (L / g). Convert the length to meters, and use 9.8 for gravity. The answer will be in seconds.
About 1.4 seconds.
A 45cm simple pendulum on earth has a period of 1.35 seconds.
2 seconds
You can use a simple pendulum, measure how long one period takes, then use the formula for a pendulum, and solve for gravitational acceleration.
You can build a simple pendulum - one that has most of its mass concentrated in a small place, at the end of the pendulum. Measure the pendulum's length, and measure how long it takes to go back and forth. Use the formula for the period of a pendulum, solving for "g".
The period of a 0.85 meter long pendulum is 1.79 seconds.
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It is assumed that the question asks "for the same period". The period of a simple pendulum, for "very short swings", is 2 pi (L/G)0.5. Since G on the moon is 0.165 that of Earth, L would have to be 0.165, so that 1 m pendulum would have to be about 0.165 m long in order to give the same swing period.
Nice problem! I get 32.1 centimeters.
You can use a simple pendulum, measure how long one period takes, then use the formula for a pendulum, and solve for gravitational acceleration.
You can build a simple pendulum - one that has most of its mass concentrated in a small place, at the end of the pendulum. Measure the pendulum's length, and measure how long it takes to go back and forth. Use the formula for the period of a pendulum, solving for "g".
ts period will become sqrt(2) times as long.
The period of a 0.85 meter long pendulum is 1.79 seconds.
It would tend towards infinity
In a freely falling elevator, there would be a period of infinite length, because in freefall, the objects act as if there is no gravity. On earth, the period is given as the quantity 2pi times the square root of the quantity length/g. g is the gravitational constant, which is 9.8 on earth. The period of the pendulum does not depend on how far you pull it back, or how much mass is on the pendulum. Both are common misconceptions. I don't have access to a calculator now, but I will come back to add the actual answer later. Use the formula within the explanation.
none. when there is gravity T=2pi square root of L/g but in a freely falling elevator, there is no accelerate so it doesn't have period the answer is none
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Its mass (weight) can be made anything you want. As long as the bob weighs significantly more than the string that suspends it, and as long as air resistance can be ignored, nothing you do to the bob has any effect on the period of the pendulum's oscillation.
It is assumed that the question asks "for the same period". The period of a simple pendulum, for "very short swings", is 2 pi (L/G)0.5. Since G on the moon is 0.165 that of Earth, L would have to be 0.165, so that 1 m pendulum would have to be about 0.165 m long in order to give the same swing period.
Answer #1:Your question cannot be answered without knowing what the pendulum wasfilled with before it was filled with mercury.If it had nothing in it, before, then adding the mercury would increase theperiod time.If it had lead in it before, then adding the mercury would decrease the periodtime.================================Answer #2:The period of a simple pendulum doesn't depend on the weight (mass) of thebob. As long as the bob is much heavier than the string, and air resistance canbe ignored, nothing you do to the bob has any effect on the period.