Describing the motion of a mass in a pendulum with an additional constant upwards force

Question

If I have a mass on a string that follows SHM and I add a constant force in the upwards y-direction to the mass, how does its period change? We assume the upwards force is less than the weight.
I had one idea, that is to sum the vertical forces, divide by g and calculate the period as for a pendulum with a mass $m_2$, such that $m_2g=mg-F$, but I am not sure whether that is correct.

smejak · Accepted Answer

This is what I came up with.
Assuming that the angle at which the mass is displaced is small enough and we neglect all forces other than gravity and the constant upwards force acting on the mass, essentially what the upwards force is doing is making the gravitational force weaker.
$sum{F_y}=F_g-F_{up}$
The net force in the y-direction is still downwards (assuming $F_g>F_{up}$), so we need to find a $g_2$ such that
$m g_2=sum{F_y}$
then we just use the formula for period of a simple pendulum to find the new period.
$T=2pi{sqrt{frac{l}{g_2}}}$

Bhavay · Answer

Hint: Calculate the new tension(T') in the mean position and equate $T'sin(theta) approx T'theta = ma$ .Now use $a=(-)omega^2 x$ to calculate the time period.

Describing the motion of a mass in a pendulum with an additional constant upwards force

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