# Are nodes on a string perfect nulls?

1. Oct 20, 2015

### musichascolors

Maybe I'm thinking about this incorrectly, but I would assume that the wave loses energy after bouncing back and before hitting the wave moving in the opposite direction. Wouldn't this prevent a perfect null (if the amplitudes were different)?

2. Oct 20, 2015

### Simon Bridge

Welcome to PF;
The short answer is "no" - there is no such thing as perfection in real life.
You are correct - energy losses, which occur throughout the string, not just at reflection, mean that the relations you learn for standing waves are only approximate.
You can imagine the situation where 100% of the wave gets transmitted at the end - in which case there is no reflection, and thus no cancellation, and so there are no nodes at all.

3. Oct 20, 2015

### musichascolors

Thanks, could you please rephrase this? No sure what you mean.

4. Oct 20, 2015

### Simon Bridge

In your discussion post #1 you imagined some energy loss after reflection - instead, imagine the energy loss occurring at the point of reflection: then we can keep ideal strings in between the reflection points.
Loss of energy at the reflection point would be equivalent to having some transmittion (along an ideal string) at that point.

The usual treatment has 100% reflection at the ends of the string of a travelling wave amplitude A - this would be the same as 0% energy loss at each reflection.
The result is a standing wave with 0 amplitude at the nodes and amplitude 2A at the antinodes.

Imagine there were 100% energy loss at the ends. Then there would be 0% reflection ... it's the same wavelength so the theoretical nodes and antinodes are in the same place, but there can be no standing wave without the reflected wave ... so, logically, what is the amplitude of the motion at the node and antinode locations?

5. Oct 20, 2015

### musichascolors

Yes, I understand that their is energy loss both before, during, and after the reflection. Thank you for clarifying that null points aren't "perfect nulls" which makes sense.