# Spring pendulum, Kinetic Energy

In the included picture, I don't get how they get to the kinetic energy part. I would say, that the travelled distance is equal to (l + x(t))*θ. Then I would take the time derivative, resulting in dx(t)/dt * θ + (l + x(t))* dθ/dt. Then I would square this result and multiply that with 1/2 m. But then I would get a totally different kinetic energy.

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mfb
Mentor
I would say, that the travelled distance is equal to (l + x(t))*θ
What does that mean? The travelled distance depends on the trajectory. You'll need a position (as vector) to get a meaningful derivative.

Vishwaas
You must also consider the radial component. The velocity component in the radial direction is the time derivative of x(t) (since it is a spring, it can compress or expand). Hence the total kinetic energy is the sum of energies in both radial and angular directions.

vanhees71
Gold Member
2019 Award
In polar coordinates, with the polar angle relative to the direction poiinting downwards, you have
$$\vec{x}=r \begin{pmatrix} \sin \varphi \\ \cos \varphi \end{pmatrix}.$$
Then, after straight-forward algebra, you have
$$T=\frac{m}{2} \dot{\vec{x}}^2=\frac{m}{2} (\dot{r}^2+r^2 \dot{\varphi}^2).$$
Now in the textbook, they set
$$r=l+x,$$
where ##l## is the length of the relaxed spring.

For the potential energy you have the part from the gravitational field of the earth and the spring:
$$V=-m g x_2+\frac{k}{2} x^2 = -m g (x+l) \cos \varphi+\frac{k}{2} x^2.$$