Frequency of a simple pendulum

[bacon]

A simple pendulum of length L and mass m is suspended in a car that is traveling with constant speed v around a circle of radius R. If the pendulum undergoes small oscillations in a radial direction about its equilibrium position, what will its frequency of oscillation be?

I don’t know how to post diagrams, so...
The negative x direction is to the left, toward the center of the circle of radius R. \theta is the angle L makes with the vertical.
The forces on m are the gravitational force and the tension along length L. The forces result in SHM for small values of \theta.

F_{x}=-kx=-mgsin\theta -m\frac{v^{2}}{R}sin\theta

For small values of \theta, \theta \approx sin\theta.
s=L\theta. Where s is the arc length L sweeps through angle \theta.
For small values of \theta, s \approx x

Using these two approximations, the top equation becomes
kx=mg\frac{x}{L} +m\frac{v^{2}x}{RL}
After cancelling the x's and rearranging, I get
\frac{k}{m}=\frac{g+\frac{v^{2}}{R}}{L}
For frequency f and period T, f=\frac{1}{T}, T=2 \pi \sqrt{\frac{m}{k}},
f=\frac{1}{2 \pi}\sqrt{\frac{g+ \frac{v^{2}}{R}}{L}}
Unfortunately, the answer in the back of the book is
f=\frac{1}{2 \pi}\sqrt{\frac{\sqrt{g^{2}+\frac{v^{4}}{R^{2}}}}{L}}

Thanks for any help.

 

[ehild]

The pendulum will oscillate around an angle θ₀ because of the centrifugal force acting in the car as rotating frame of reference: tan(θ₀)=v²/(Rg). The deviation from this angle can be considered small, but not the angle with respect to vertical.

ehild