The acceleration of free fall on the moon is one-sixth of that on the earth. What would be the period on the moon of a simple pendulum which has a period of one second on the earth?
Correct answer: A. √6 sec
- A. √6 sec
- B. 6 s
- C. 1 /√6 s
- D. 6/ √6 s
Explanation
The period T of a simple pendulum is given by the formula T=2π(L/g)^1/2 , where L is the length of the pendulum and g is the acceleration due to gravity. On the moon, the acceleration due to gravity (gm ) is one-sixth of that on the earth (ge ). Therefore, Tm =2π( L/gm)^1/2 = 2π( L/1/6(ge))^1/2 =√6 ×2π( L/ge)^1/2 =6 ×Te =√6x1 sec= √6 sec
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About Simple Harmonic Motion
Simple harmonic motion is oscillation in which acceleration is directly proportional to displacement and directed toward the equilibrium position. Work includes displacement, velocity, acceleration, phase, period, frequency, amplitude and energy, with applications to springs and simple pendulums. The restoring force and the conditions for SHM distinguish it from general periodic motion.
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