A wave on a string travels at a speed set by the string. What happens when it reaches the end, or a knot joining a different string? Part of it bounces back, and how it bounces depends on the boundary.
Key idea
A pulse reflects upside down from a fixed end (or from a heavier, slower string) and right side up from a free end (or from a lighter, faster string). At a junction, part of the pulse also passes through, always right side up, at the new speed. The frequency never changes at a boundary.
Why it works
At a fixed end, the arriving pulse pulls up on the wall. By Newton's third law the wall pulls down on the string, and that downward pull launches an upside-down pulse back. At a free end (a ring sliding on a pole), nothing holds the end, so it swings up even higher and launches an upright pulse. A heavier string acts partly like a wall; a lighter one acts partly like a free end.
Common mistake
Thinking a reflected pulse always comes back the same way up, like a ball bouncing off a wall. A pulse is a shape moving along a string, not an object. Check: at a fixed end the end itself never moves, which only works if the incoming and reflected pulses cancel there, so the reflected one must be inverted.
Worked example: A pulse round trip
A pulse moves at $40\ \text{m/s}$ on a $6.0\ \text{m}$ string tied to a wall. It starts upright at your hand. How long until it gets back to you, and which way up is it?
- It travels to the wall and back: $2 \times 6.0 = 12\ \text{m}$.
- $t = \dfrac{12\ \text{m}}{40\ \text{m/s}} = 0.30\ \text{s}$.
- The wall is a fixed end, so the pulse comes back upside down.
Answer: $0.30\ \text{s}$, upside down
Worked example: Into a heavier string
A pulse $20\ \text{cm}$ wide moves at $30\ \text{m/s}$ on a light string tied to a heavier string, where waves move at $15\ \text{m/s}$. Describe the reflected and transmitted pulses.
- Reflected: a heavier string acts partly like a wall, so the reflected pulse is upside down, still $30\ \text{m/s}$ and $20\ \text{cm}$ wide.
- Transmitted: right side up, moving at $15\ \text{m/s}$.
- The front of the pulse crosses the knot first and slows down, so the back catches up: the width shrinks in proportion to the speed, $20 \times \frac{15}{30} = 10\ \text{cm}$.
Answer: Reflected inverted and $20\ \text{cm}$ wide; transmitted upright, $10\ \text{cm}$ wide
Light does the same thing. Reflecting off a material with a higher index of refraction, where light is slower, flips the wave over. That flip is the key to thin-film colors later in this unit.