A 50 cm long mathematical pendulum was placed in an elevator. Determine the oscillation frequency of this pendulum if the lift moves upward with an acceleration of 1.2 m / s2.
Let’s translate the values from given to the SI system:
l = 50 cm = 0.5 m.
The oscillation period is such a period of time during which the body returns to the same point from which it began its movement.
T = 2π * √ (l / g)
In our case, the direction of acceleration of the body and the acceleration of gravity are oppositely directed and in the same plane, then the expression takes the form:
T = 2π * √ (l / (g + a))
The oscillation period depends on the frequency according to the law:
T = 1 / ν
Let us express the frequency from this expression:
ν = 1 / T
Substitute the formula for the period into this expression:
ν = 1 / T = 1 / (2π * √ (l / (g + a)))
substitute the numerical values and determine the frequency:
ν = 1 / (2π * √ (l / g + a)) = 1 / (2π * √ (0.5 / (9.8 + 1.2) -)) = 0.712 Hz.
Answer: the oscillation frequency is 0.712 Hz.
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