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The frequency of oscillation of a simple pendulum of length L, mounted in a cabin that is falling freely under gravity is
- a)infinity
- b)zero
- c)√(g/2L)
- d)√(g/L)
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The frequency of oscillation of a simple pendulum of length L, mounted...
Option B is correct
explanation
The answer is zero - the pendulum will not be in oscillation.
Even if the pendulum had been in oscillation before the commencement of the cabin’s free fall, this oscillation will cease very quickly once the free fall begins.
Hitherto, the pendulum swing was due to three identifiable forces: 1) The downwardly directed force of gravity 2) The tension in the string 3) The angular momentum of the pendulum.
At the commencement of the cabin’s free fall, whereas the downwardly directed force of gravity still acts on the pendulum, the tension force in the string disappears because the cabin roof, presumably from which the pendulum hangs, is in free fall also and thus a “reaction” pull by the string becomes inexistent since there is an inexistent pull by the pendulum mass also, as both objects have zero motion relative to each other, nor is there a dynamic equilibrium of forces acting between them.
The angular momentum earlier, was a resultant of the gravitational pull on the pendulum against the ever present tension on the string (drivable by the parallelogram of forces theorem and ever changing its direction as a new component, with every new component remaining perpendicular to the tension force in the string at any point in time along the swing).
With the downward gravitational pull and the tension in string effectively neutralized by the free fall of the cabin, angular momentum also ceases and with it any further swinging motion of the pendulum.
explanation
The answer is zero - the pendulum will not be in oscillation.
Even if the pendulum had been in oscillation before the commencement of the cabin’s free fall, this oscillation will cease very quickly once the free fall begins.
Hitherto, the pendulum swing was due to three identifiable forces: 1) The downwardly directed force of gravity 2) The tension in the string 3) The angular momentum of the pendulum.
At the commencement of the cabin’s free fall, whereas the downwardly directed force of gravity still acts on the pendulum, the tension force in the string disappears because the cabin roof, presumably from which the pendulum hangs, is in free fall also and thus a “reaction” pull by the string becomes inexistent since there is an inexistent pull by the pendulum mass also, as both objects have zero motion relative to each other, nor is there a dynamic equilibrium of forces acting between them.
The angular momentum earlier, was a resultant of the gravitational pull on the pendulum against the ever present tension on the string (drivable by the parallelogram of forces theorem and ever changing its direction as a new component, with every new component remaining perpendicular to the tension force in the string at any point in time along the swing).
With the downward gravitational pull and the tension in string effectively neutralized by the free fall of the cabin, angular momentum also ceases and with it any further swinging motion of the pendulum.
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The frequency of oscillation of a simple pendulum of length L, mounted in a cabin that is falling freely under gravity isa)infinityb)zeroc)√(g/2L)d)√(g/L)Correct answer is option 'B'. Can you explain this answer?
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