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Identical balls oscillate with the same period T on Earth. Ball A is attached to an ideal spring and ball B swings back and forth to form a simple pendulum. These systems are now taken to the Moon, where g = 1.6 m/s2, and set into oscillation. Which of the following statements about these systems are true? (There could be more than one correct choice.)

A) Both systems will have the same period on the Moon as on Earth.
B) On the Moon, ball A will take longer to complete one cycle than ball B.
C) On the Moon, ball B will take longer to complete one cycle than ball A.
D) On the Moon, ball A will execute more vibrations each minute than ball B.
E) On the Moon, ball B will execute more vibrations each minute than ball A.

User Tonytran
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3.7k points

2 Answers

3 votes

Answer:

C , D

Step-by-step explanation:

The time period of loaded spring is given by


T=2\pi \sqrt{(m)/(K)}

where, m is teh mass of the body and K is the spring constant

The time period of the simple pendulum is given by


T' = 2\pi \sqrt{(l)/(g)}

where, l is the length of the pendulum and g be the acceleration due to gravity

So, the time period of A does not matter as it is independent of g, but the time period of B is more than earth as it reaches on moon.

As teh spring

User Erdi Aker
by
3.2k points
7 votes

Answer:

Step-by-step explanation:

Time period of a spring mass system does not depend upon gravitational acceleration ( g ) , so time period of ball A will remain unchanged.

Time period of a pendulum is inversely proportional to g so time period of ball B will be increased or its oscillation will become slow.

option C and D are correct.

User Bahri Gungor
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3.2k points