Can KE and PE of a small element of a transverse progressive wave be maximum simultaneously?

Question

Maybe a string can be taken as an example to produce the transverse progressive wave.

Answer 1

Let one end of a very long string is being oscillated transversely so as to generate a sinusoidal wave traveling out along the string.

In order to set up a wave on a stretched string, the driving force at the end of the string provides energy. This energy is not retained at the source; it flows along the string at the wave speed.

The string transports energy as both kinetic energy & elastic potential energy.

Kinetic Energy:

As the wave passes through a small string element, it oscillates transversely in SHM. When the element is rushing through its $y = 0$ position, its transverse velocity - thus kinetic energy - is maximum as evident from the formula $KE = \frac{1}{2} \mu dx {\left(\dfrac{dy}{dt} \right)}^2$ where $y = A \sin{\left[\dfrac{2\pi}{\lambda} (x - vt) \right]}$. When the element is at its extreme position $y = A$, then the kinetic energy is zero. All the associated kinetic energy by then has been trnsferred to the next segment of the string as that part has now come to $y = 0$.

Elastic potential energy:

To send a sinusoidal wave along a previously straight string, the wave must stretch the string. As a string of length $dx$ oscillates transversely, its length must increase & decrease in a periodic way if the string element is to fit the sinusoidal form.

When the string element is at its $y = A$, its length is normal undisturbed value $dx$. However, when the element is rushing through its $y = 0$, it has maximum stretch & thus maximum elastic potential energy.

Thus the oscillating string element has both its maximum kinetic energy & maximum elastic potential energy simultaneously at $y = 0$."-Principles of Physics; Extended 9th edition by Walker, Resnick, Halliday.

_{(source: cnx.org)}

Answer 2

No. Energy conservation always applies. The elastic potential energy will be maximum at a wavetop, since here the rope is stretched the most, $U=½kx^2$. The transverse velocity and thus the kinetic energy is zero at this point $K=½mv^2$ since this part of the rope stops and starts moving back again.

$$E_{before}=E_{after} \implies K_1+U_1=K_2+U_2$$

Energy before will still be energy after.