We have two oscillators, used to measure time. One is a vertical mass-and-spring system. The other is a simple pendulum. On earth the two systems have the same period. How do their periods compare on the moon? A. Both periods become longer. B. only the pendulum\'s period becomes longer C. Only the mass-and-spring\'s period becomes longer D. Neither period changes at all. E. none of the above If a mass hangs from a vertical spring and is moving up and down, you take one video of the motion on earth and then take another video of its motion on the moon, can you tell the difference between the two motions? Solution time period of simple pendulam= 2*pi sqrt(l/g) so time period inversly propotional to gravitaion constant. time period of verticle ossilator= 2*pi sqrt(k/m) so time period does not depend on g. we knows g of moon < g of earth. so on moon g is less then that of earth. so g decreases, so time period of pendulam increase as time period of pendulan is inverse of sqrt g. and time period of oscillator remains the same. so option B. only pendulam\'s timeperiod becomes longer. .

1. We have two oscillators, used to measure time. One is a vertical mass-and-spring system. The
other is a simple pendulum. On earth the two systems have the same period. How do their
periods compare on the moon?
A. Both periods become longer.
B. only the pendulum's period becomes longer
C. Only the mass-and-spring's period becomes longer
D. Neither period changes at all.
E. none of the above
If a mass hangs from a vertical spring and is moving up and down, you take one video of the
motion on earth and then take another video of its motion on the moon, can you tell the
difference between the two motions?
Solution
time period of simple pendulam= 2*pi sqrt(l/g)
so time period inversly propotional to gravitaion constant.
time period of verticle ossilator= 2*pi sqrt(k/m)
so time period does not depend on g.
we knows g of moon < g of earth.
so on moon g is less then that of earth.
so g decreases, so time period of pendulam increase as time period of pendulan is inverse of sqrt
g.
and time period of oscillator remains the same.

2. so option B. only pendulam's timeperiod becomes longer.