Simple Harmonic Motion

[jonnburton]

Hi,

I am trying to learn about simple harmonic motion/oscillation and am looking at the energy of the system.

The kinetic energy is \(\displaystyle \frac{1}{2}mv^2 = \frac{1}{2}m\dot{x}^2\) and the potential energy is \(\displaystyle \frac{1}{2}kx^2\)

Since the energy is conserved,

\(\displaystyle \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2 = constant\)


I don't understand the next step outlined in the book: obtaining the differential equation describing the accelerated motion of the block.

It says this can be obtained by differentiating the equation with respect to time, giving:

\(\displaystyle m\ddot{x}\dot{x} + kx\dot{x}=0\)


My problem is I can't see how to differentiate the equation \(\displaystyle \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2 = constant\).

Putting it in notation with which I am more familiar (even though it means the same as the above):

\(\displaystyle \frac{1}{2}m(\frac{dx}{dt})^2 + \frac{1}{2}kx^2 = constant\)

and just focussing on the first part: \(\displaystyle \frac{1}{2}m(\frac{dx}{dt})^2\)

This has to be differentiated with respect to t:

\(\displaystyle \frac{d}{dt} (\frac{1}{2}m(\frac{dx}{dt})^2)\)


Surely this has to be done using the product rule, which when I do it, yields:

\(\displaystyle \frac{1}{2}m * 2\frac{dx}{dt} + (\frac{dx}{dt})^2\)

\(\displaystyle = m\frac{dx}{dt} + (\frac{dx}{dt})^2\)

\(\displaystyle = m\dot{x} + \dot{x}^2\) and not \(\displaystyle m\ddot{x}\dot{x}\)



Can anybody tell me where I am going wrong?

 

[DrPhil]

Hi,

I am trying to learn about simple harmonic motion/oscillation and am looking at the energy of the system.

The kinetic energy is \(\displaystyle \frac{1}{2}mv^2 = \frac{1}{2}m\dot{x}^2\) and the potential energy is \(\displaystyle \frac{1}{2}kx^2\)

Since the energy is conserved,

\(\displaystyle \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2 = constant\)


I don't understand the next step outlined in the book: obtaining the differential equation describing the accelerated motion of the block.

It says this can be obtained by differentiating the equation with respect to time, giving:

\(\displaystyle m\ddot{x}\dot{x} + kx\dot{x}=0\)


My problem is I can't see how to differentiate the equation \(\displaystyle \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2 = constant\).

Putting it in notation with which I am more familiar (even though it means the same as the above):

\(\displaystyle \frac{1}{2}m(\frac{dx}{dt})^2 + \frac{1}{2}kx^2 = constant\)

and just focussing on the first part: \(\displaystyle \frac{1}{2}m(\frac{dx}{dt})^2\)

This has to be differentiated with respect to t:

\(\displaystyle \frac{d}{dt} (\frac{1}{2}m(\frac{dx}{dt})^2)\)


Surely this has to be done using the product rule, which when I do it, yields:

\(\displaystyle \frac{1}{2}m * 2\frac{dx}{dt} + (\frac{dx}{dt})^2\)...This is wrong .. see below

\(\displaystyle = m\frac{dx}{dt} + (\frac{dx}{dt})^2\)

\(\displaystyle = m\dot{x} + \dot{x}^2\) and not \(\displaystyle m\ddot{x}\dot{x}\)



Can anybody tell me where I am going wrong?

\(\displaystyle \displaystyle\frac{d}{dt} \left(\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 \right) \)

NOT a product .. think "power rule and chain rule" instead. Remember that \(\displaystyle \frac{d}{dt}\) is an operator, not a multiplier.

\(\displaystyle \displaystyle \frac{d}{dt}\left( \dot{x}^2\right) = 2\ \dot x\ \frac{d}{dt}\dot x = 2\ \dot x\ \ddot x\)

because the \(\displaystyle \frac{d}{dt}\) operating on a first derivative gives the 2nd derivative.

 

[jonnburton]

Thanks a lot DRPhil!

\(\displaystyle \displaystyle\frac{d}{dt} \left(\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 \right) \)

NOT a product .. think "power rule and chain rule" instead. Remember that \(\displaystyle \frac{d}{dt}\) is an operator, not a multiplier.

\(\displaystyle \displaystyle \frac{d}{dt}\left( \dot{x}^2\right) = 2\ \dot x\ \frac{d}{dt}\dot x = 2\ \dot x\ \ddot x\)

because the \(\displaystyle \frac{d}{dt}\) operating on a first derivative gives the 2nd derivative.