My question pertains to the following excerpt from my lecture notes:

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Part (b) of figure 2.12 shows a cartoon of a snapshot some time later. Both transmitted and reflected pulses have a smaller amplitude that the incident pulse, and the transmitted pulse has travelled a shorter distance in the intervening time as its speed is lower. The difference in wave speed in the two sections means that the transmitted pulse appears compressed along the x-axis. The wave pulse is ultimately caused by the source of the wave, so its duration (i.e., length in time) must be fixed. If the disturbance lasts for the same length of time, but its speed is reduced, then it must be narrower in space.

I have trouble interpreting why the transmitted wave is narrower in space.

I don't understand "the wave pulse is ultimately caused by the source of the wave, so its duration must be fixed". Yes, it's caused by the source of the wave, but where does this then imply the duration is fixed?

Thus, it is stated "If the disturbance lasts for the same length of time, but its speed is reduced, then it must be narrower in space." However, I don't see how these two points imply the third.

Can someone help in illuminating this for me?

  • 2
    $\begingroup$ The picture is Figure 2.13, but the copied text refers to Figure 2.12. $\endgroup$ – Mark H Mar 6 '18 at 23:56

The duration of the wave is fixed as it travels along the string because the speed of each part of the wave profile is the same at every point along the string. This happens because the string is assumed to be a linear medium. The high-frequency parts of the wave (the rising and falling edges at half amplitude, where the displacement is changing fastest) move along the string at exactly the same speed as the low-frequency parts (the peak and the 'flat' leading and trailing sections, where displacement is changing slowest). In technical terms, there is no dispersion.

The profile of the wave in the time domain (displacement vs time at a fixed point on the string) remains the same along the string, even when in regions where wave speed is different. The time delay between any two reference points on the wave passing a fixed point on the string is the same at all points along the string. The 2 reference points cross the boundary and enter the slower region separated by the same time delay, so they retain the same separation in the slower region. Hence the pulse duration is the same in the slower region.

However, in the space domain (displacement vs distance along the string at a fixed instant in time), which is used in the diagram in your question, the length of the wave can change. The wave-length and speed change in proportion, the constant of proportionality being the duration of the wave.

As an analogy, imagine identical soldiers being 'emitted' from their HQ at regular intervals of 5s (ie at constant frequency). When marching along a road at 2m/s they are a constant distance of 10m apart. When they move onto rough ground they slow down to 1m/s and bunch up. Each soldier crosses the boundary 5s ahead of the one behind him, but he has only advanced 5m from the boundary when the next one crosses it. The distance between soldiers (wavelength) falls in proportion with their speed, but the time separation remains 5s regardless of the terrain.

If the high-frequency sections of the wave (steep edges) travel faster than the low-frequency sections (peak and trough) then the pulse gets shorter in the time domain. In this case there is dispersion. This happens when the medium is non-linear. An example is water waves : high-frequency capillary waves caused by surface tension travel faster than low-frequency gravity waves, so the ripple spreads out as it emanates from the origin.

Returning to the army analogy, imagine 2 soldiers are 'emitted' from HQ 50s apart. The 2nd is taller; he marches at the same pace but because his stride is longer he marches at 4m/s along the road compared with 2m/s for the shorter man. When the 2nd soldier sets out they are 100m apart. After another 20s the 2nd soldier has moved closer to the 1st. The 1st is now 140m from HQ, but the 2nd is 80m from HQ. They are now only 60m apart. Not only are they getting closer in the space domain, they are also getting closer in the time domain. Whereas the 2nd soldier left HQ 50s behind the 1st, and reaches the 100m point 100/4=25s behind the 1st, he reaches the 140m point only 60/4 = 15s behind.

  • $\begingroup$ When different parts of a wave move at different speeds, there is dispersion. For example water waves. $\endgroup$ – Pieter Mar 7 '18 at 7:53
  • $\begingroup$ So why is the duration of the wave pulse in the string with $v_2$ the same as in the string with $v_1$? $\endgroup$ – freecharly Mar 7 '18 at 20:10

When the wave hits the boundary, it starts to slow down because the density of the second portion of rope is higher than the first. Slowing down the wave has the effect of squishing the peak. You can intuit this by considering the motion of the two tails of the wave packet.

Consider the forward tail as it hits the boundary, it begins to travel at the speed of a wave in the new material, it slows down due to the increased density and constant tension. While it advances forward, the trailing tail is traveling toward the boundary at the faster speed of the first portion of the rope. While the wave is split between two different portions of the rope, the forward tail is travelling at the slower speed and the trailing tail is gaining on it. Once both ends of the wave-packet have made it into the slower portion of the rope, the width of the packet remains fixed.


A wave propagating in positive x-direction satisfying the standard wave equation with wave velocity $v_1$ has the general functional form $$f(x-v_1t) \tag 1$$ where $f$ is an arbitrary function of its argument, and a wave propagating in negative x-direction has the general form $$g(x+v_1t) \tag 2$$ where $g$ is an arbitrary function of its argument. Assuming that at a fixed position $x_1$ the wave deviates from zero for a time $\tau_1=t_b-t_a$ in medium 1 with wave velocity $v_1$, you can easily see from finding the x-values $x_a$ and $x_b$ for $t=t_a, t_b$ where $f(x-v_1 t)= 0$ that the length of the wave pulse in medium 1 is always $l_1=v_1 \tau_1$. The pulse duration $\tau_1$ of the incident wave pulse $f_i$ also holds at the transition point $x_0$ to the medium 2 with a different wave velocity $v_2$. At this point, the amplitude of the incident wave is $$f_i(x_0-v_1 t) \tag 3$$ the amplitude of the reflected wave is $$f_r(x_0+v_1t)=f_i(x_0-v_1t)\cdot R \tag 4$$ the amplitude of the transmitted wave is $$f_t(x_0-v_2t)=f_i(x_0-v_1t)\cdot T\tag 5$$ The factors $R=\frac {Z_2-Z_1}{Z_2+Z_1}$ and $T=\frac {2Z_2}{Z_2+Z_1}$ are the reflection and transmission coefficients, where $Z_1$ and $Z_2$ are the characteristic wave impedances of medium 1 and medium 2, respectively. From this follows that the time duration of the incident, the reflected and the transmitted wave pulses is always $\tau_1$. The spatial length of the incident and reflected wave pulses in medium 1 is always $l_1=v_1 \tau_1$ and the spatial length of transmitted wave pulse in medium 2 is $l_2=v_2 \tau_1$. One can also see that the wave forms of the reflected and transmitted waves are just scaled functions of the incident wave function.

  • $\begingroup$ The use of lower-case $r$ and $t$ for the coefficients is confusing because $t$ is already used to denote time. Still, I believe this answer more correctly addresses the question than the accepted answer because it shows how the linear period, $\tau$ does not change across media. $\endgroup$ – D. Betchkal Mar 8 '18 at 2:11
  • $\begingroup$ @D.Betchkal Thank you, I will change the coefficients to capital letters. $\endgroup$ – freecharly Mar 8 '18 at 2:47

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