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

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

• I asked the same question a year ago but also got downvoted. However, I'll post an answer to your question & will show that your statement is true; but it doesn't violate SHM.
– user36790
May 20, 2015 at 6:07
– Andy
May 20, 2015 at 6:10
• Okay. But I don't understand how we both are correct. Maybe you could answer that too. Thanks.
– Andy
May 20, 2015 at 6:14
• I believe that the downvote is for a very short and broad question. Also, you do not show why you would think that PE and KE should both be maximum. May 20, 2015 at 6:47
• Steeven and user36790 are both correct, since they both define PE differently. OP should've mentioned which PE he/she is referring to.
– Sidd
May 20, 2015 at 10:18

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)

• I've added the answer as promised. You can check this also. Hope this helps.
– user36790
May 20, 2015 at 9:46

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.

• So the element at the top will be the one having max PE?
– Andy
May 20, 2015 at 6:15
• @Andy Yes. Elastic (and gravitational, if the wave is vertical) potential energy is max for that point. At the same time, the kinetic energy is zero as the particle stops and gets ready to move back. May 20, 2015 at 6:40