# Kinetic energy stored in a triangular wave

I usually used to solve questions based on plane progressive harmonic wave. Recently I encountered a triangular plane progressive wave. Here is a snapshot of the wave.

I am provided with mass/unit length of the string and the tension in the string. I had to find out the kinetic energy of the wave pulse travelling in the taut string as in the figure above.

Here is my attempt:

The triangular wave seems to me like $\sin^{-1}(\sin x)$ with some modifications. I am struggling finding out correct expression of $y$ as expression of $x$ and then integrate to find out the required energy. Any help is appreciated.

Outline of solution:

• Find the total amount of potential energy stored. This is simply $T \Delta \ell$ where $T$ is the tension and $\Delta \ell$ is the change in length of the rope.
• Don't bother doing integrals, just compute $\Delta \ell$ with the Pythagorean theorem.
• Note that for waves on a string, the kinetic and potential energies are equal.
• Thanks for your answer. That means we donot require mass per unit length of the string. – Pink Apr 18 '17 at 6:18
• Could you explain why kinetic and potential energies are equal ? – user139621 Apr 18 '17 at 6:37
• @knzhou Please have a look at my answer. – Farcher Apr 18 '17 at 9:36
• It is very difficult to find $\Delta L$ – Koolman Apr 18 '17 at 12:05

I assume that this is a pulse going along the string and as such the profile of the pulse does not change with time.

In the diagram the pulse is shown at time $t=0$ (in grey) and then at a later time $\tau$ (in red) where $c$ is the speed of the wave.

The potential energy stored in the pulse does not change but the wave profile as shown by the OP does not include the fact that the string is moving as shown with the green arrows in my diagram.
That is where the kinetic energy of the pulse is and it is constant as the pulse moves along.

To maintain the wave profile the downward velocity of the tailing edge of the pulse must be the same all along that part of the string and the different upward velocity of the pulse must be the same all along the leading edge.

The kinetic energy of the pulse can be found by relating the upward and downward velocities of the string to the horizontal speed of the pulse $c$ which will be related to the mass per unit length of the string and the tension in the string.

Update as a result of a comment

The displacement and velocity triangles are similar.

So the speed of the le left hand side of the string is $v = \dfrac {c}{100}$

So the kinetic energy of the left hand side of the string is $\dfrac 1 2 \mu \cdot 100 \cdot \left ( \dfrac {c}{100}\right )^2 = \dfrac 1 2 T \dfrac{1}{100}= \dfrac{T}{200}$ whwere $\mu$ is the mass per unit length of the string and $\mu = \dfrac{T}{c^2}$ with $T$ as the tension in the string.

If the extension of the left hand side of the string is $x$ then $(100+x)^2 = 100^2+1^2 \Rightarrow x = \dfrac {1}{200}$ if one discards the very small $x^2$ term.

The kinetic energy of the left hand side of the string is $Fx$ and adding an equivalent amount of kinetic energy for the right hand side of the string gives @knzhou 's value for the potential energy in terms of the tension and the extension of the string.

• Thanks for your answer. The horizontal speed of the pulse can be related as $v=\sqrt{\frac{T}{\mu}}$. But i have no idea on how to relate with it upward and downward velocity. My idea is that it is related to slope of curve as $v'=-v(slope)$ as it is the particle velocity in a harmonic wave. Can you please elaborate on it. Thanks. – Pink Apr 18 '17 at 11:49
• Some geometry and trigonometry should help you? You have an angle(s) and a side $c\tau$ or $c$ of a right angled triangle and so you should be able to find one of the other sides? – Farcher Apr 18 '17 at 11:55
• Very nice. Thanks. But i how shall i relate this to horizontal velocity. – Pink Apr 18 '17 at 12:03
• Looking at my diagram, draw a right angle triangle with $c$ or c\tau\$ as the hypotenuse. – Farcher Apr 18 '17 at 12:06
• @navinstudent This answer will give you the exact same result as mine; it's basically a proof that the kinetic and potential energies are equal. – knzhou Apr 18 '17 at 16:12