# If taken to the Moon, will there be any change in the frequency of oscillation of a oscillator? A physical pendulum?

Jitender Pal
9 years ago
For a torsional pendulum, frequency of oscillation (f) is defined as,
f = 1/2π √k/I
Here k is the force constant and I is the rotational inertia of the body.
For moon gʹ = g/6. But the frequency is independent of g. It depends on k and I, which are always constant for a given mass and cord. So the frequency does not change.
For simple pendulum,
f = 1/2π √g/l
Here g is the acceleration due to gravity and l is length of pendulum. For moon gʹ = g/6. So the frequency of oscillation depends up on acceleration due to gravity g. It changes when you take the simple pendulum to moon due to gravitational acceleration and it becomes,
f = 1/2π √g/6l.
For simple block oscillator, frequency of oscillation (f) is defined as,
f = 1/2π √k/m
Here k is the force constant and m is the mass of block.
From the above equation f = 1/2π √k/m, we observed that the frequency of oscillation f is independent of acceleration due to gravity g. Therefore it also does not change.
For physical pendulum, frequency of oscillation (f) is defined as,
f = 1/2π √Mgd/I
Here M is the mass of pendulum, g is the acceleration due to gravity, d is the distance from the pivot to the center of mass and I is the rotational inertia of body.
Since the frequency of oscillation depends up on g, therefore it also changes and it becomes,
f = 1/2π √Mgd/6l