Let us suppose we have two identical waves traveling in opposite direction with 1.0 m/s speed. Both of the waves have same positive head and the width of the pulse is also the same.

Therefore, the speeds of the first wave and the second wave are given as

$$
begin{align*}
v_1 & = 1.0 mathrm{m/s} \
v_2 & = -1.0 mathrm{m/s}
end{align*}
$$

As time passes the first wave move towards right and the second wave move towards left.

$$
textbf{(a)}
$$

The sketch of the resultant waves at $t=1.0$ s is shown in the given figure.

In this graph, both the pulses are far from each other.

$$
textbf{(b)}
$$

The sketch of the resultant waves at $t=2.0$ s is shown in the given figure.

In this graph, both the pulses are side by side (but not meeting). Therefore, the width of the resultant pulse has increased.

$$
textbf{(c)}
$$

The sketch of the resultant waves at $t=2.5$ s is shown in the given figure.

In this graph, both the pulses are exactly meeting each other. Therefore, the height of the resultant pulse has increased but the width is the same.

$$
textbf{(d)}
$$

The sketch of the resultant waves at $t=3.0$ s is shown in the given figure.

In this graph, both the pulses are side by side (but not meeting). Therefore, the width of the resultant pulse has increased.

$$
textbf{(e)}
$$

The sketch of the resultant waves at $t=4.0$ s is shown in the given figure.

In this graph, both the pulses are far from each other.

Yes, the classmate is correct.

Let us suppose that two waves have the same amplitude and wavelength, and they interfere with each other.
If these two waves have the same phase they interfere constructively. If they have $phi = pi$ phase difference than they interfere destructively.

All the phases between zero to $2pi$ give a resultant wave with different amplitudes. Hence, we see a wave having amplitude from minimum to maximum (or maximum to minimum) at the same place. Therefore, we see a standing wave.

Hence, a standing wave involves both constructive and destructive interference which can be seen in the given graphs.

The second graph shows the interference of the two waves when phase difference is $0le phi le pi$.

The third graph shows the interference of the two waves when phase difference is $phi = pi$.

The fourth graph shows the interference of the two waves when phase difference is $pile phi le 2pi$.

The fifth graph shows the interference of the two waves when phase difference is $phi = 2pi$.

Higher the harmonic, higher the number of nodes (and so anti-nodes) and hence shorter wavelength. Since f = $dfrac{v}{wavelegth}$

we get, lower the wavelegth, higher is the frequency.

In short,yes,higher harmonics always have greater frequencies than lower harmonics

$textbf{(a)}$
A guitar makes standing waves with nodes at the ends.

In the given figure, we see that–

The fundamental mode (first harmonic) is shown which has two nodes and one antinode.

The second harmonic has three nodes and two antinodes.

The third harmonic has four nodes and three antinodes.

Therefore, the vibrating guitar with two antinodes has second harmonic.

$textbf{(b)}$ The length of the string is given as

$$
begin{align*}
L & = 66 mathrm{cm}
end{align*}
$$

As shown in the given figure, the second harmonic has wavelength equal to the length of the string.

Therefore, the wavelength of the wave is

$$
begin{align*}
lambda & = L \
lambda & = 66 mathrm{cm} \
& hspace*{-3mm}boxed{lambda = 66 mathrm{cm} }
end{align*}
$$

(a) Second harmonic.

(b) ${lambda = 66 mathrm{cm} }$

We know that when a standing wave is formed in a string the number of antinodes determines the harmonic of the wave. This gives us a result that the length of the string is equal to a integer number of wavelength halves, that is:
begin{align*}
L = n cdot frac{lambda}{2}
end{align*}
Since the length of the string is $ L = 33text{ cm}$ we get:
begin{align*}
2 cdot L = 2 cdot 33text{ cm} = 66text{ cm} = n cdot lambda
end{align*}
From this we see that any wavelength $lambda$ that fulfils this equation for an integer $n$ can produce a standing wave. \
begin{enumerate}[For]
item
$lambda_1 = 0.22text{ m}$ we get $n_1 = frac{66text{cm}}{22text{ cm}} = 3$
item
$lambda_2 = 0.33text{ m}$ we get $n_2 = frac{66text{cm}}{33text{ cm}} = 2$
item
$lambda_3 = 0.44text{ m}$ we get $n_3 = frac{66text{cm}}{44text{ cm}} = 1.5$
item
$lambda_4 = 0.66text{ m}$ we get $n_4 = frac{66text{cm}}{66text{ cm}} = 1$
end{enumerate}
Since $n_1$ , $n_2$ and $n_4$ are integers all of these wavelengths $lambda_1 , lambda_2 , lambda_4$ form standing waves. \\
These are actually the third, second and first harmonics.

We know that for the first harmonic only half a wavelength fits on a string. Therefore we can write:

$$
begin{align*}
L = frac{lambda}{2}
end{align*}
$$

Plugging in the value for the wavelength of the first harmonic $lambda = 0.65text{ m}$ we get:

$$
begin{align*}
L = frac{0.65text{ m}}{2} = 0.325text{ m}
end{align*}
$$

The distance between the walls is $L = 0.325text{ m}$