# Is there a difference in the period of a pendulum measured in inertial and noninertial frames of reference?

I was working on a pendulum in an elevator problem where the elevator was accelerating upwards, when i was working out the problem from the perspective of someone in the elevator, i found out that the forces on the bob are the components of the gravitational force mg and the inertial force ma, both of which will contribute to the period of the pendulum measured by that observer by the relation $T=2\pi (\frac L{a+g})\frac 12$. However, when i worked out the problem from the perspective of an observer in an inertial frame of reference, i found that the only force acting towards the equilibrium position was the component of the gravitational force (that's if my analysis is correct, i assumed that the mass accelerates upwards only because of the tension in the spring and otherwise would fall to the ground, and since the tension does not contribute to the forces acting towards the equilibrium position, then it does not contribute to the period) and therfore the period of the oscillation is given by $2\pi (\frac Lg)\frac 12$, which is clearly not the same as the first period. This seems paradoxical to me and i really can't find where i went wrong. Are the periods really different or is there some force that i forgot about in my analysis? Sorry if my question is stupid

• I've added the homework-and-exercises tag. In the future, please use this tag on homework problems. – Ben Crowell Mar 18 '18 at 15:21
• Clear. I didn't know these exist – Ahmed Ayman Mar 18 '18 at 15:22
• What in your analysis in the inertial frame was the tension in the string? Remember that the mass is accelerating upwards when the string is vertical. – Farcher Mar 18 '18 at 15:23
• I got the equation that T-mgcos θ=macosθ – Ahmed Ayman Mar 18 '18 at 15:28

If an object ( a pendulum ) is accelerating upwards at a rate of $a$ , then the gravitational force felt by this object is effectively, $$g_{eff}=g+a$$
In your particular case, the common equation of motion for a pendulum, $$\frac{d^2θ}{dt^2}=mgsinθ$$ replaced g with the effective g and substituted: $$\frac{d^2θ}{dt^2}=mg_{eff}sinθ=m(g+a)sinθ$$
In the inertial frame , you can think as the point towards the pendulum mass is falling is also going up with the same acceleration $a$ ( with the surrounding place).So , the pendulum mass is falling towards it with $g+a$ .