Why does a wave on a string travel at the same speed no matter how hard you pluck it?

The speed is a property of the string, not of the pluck. A harder pluck makes a taller wave that carries more energy, but the crest still advances at the same rate, because on a stretched string that rate is fixed by the tension and the mass per unit length.

Take a string 2 m long, clamped at both ends, on which disturbances travel at 10 m/s. The longest standing wave has wavelength twice the length: $\lambda_1=4$ m. Its frequency is $f_1=v/\lambda_1=10/4=2.5$ Hz. The next allowed shape has $\lambda_2=L=2$ m, so $f_2=10/2=5$ Hz. The one after has $\lambda_3=2L/3=4/3$ m $=1.333\ldots$ m, so $f_3=10/(4/3)=7.5$ Hz. Check each with $v=f\lambda$: 4 m × 2.5 Hz = 10 m/s, 2 m × 5 Hz = 10 m/s, and 1.333… m × 7.5 Hz = 10 m/s. The string has not been told which shape it is carrying and does not care.

Plucking harder does not single out a new speed. If the string is pulled aside in the same shape and released from a larger displacement, every allowed standing shape in the mixture is multiplied by the same factor. Their wavelengths stay fixed by the clamps, so their frequencies stay fixed too. What changes is how much of each shape is present, which is why a harder pluck sounds louder; it does not change the rate at which any of those shapes’ travelling counterparts move. If you pluck at a different point, the mix of shapes changes and the tone changes, but the speed of each ingredient is still set by the string.

The natural guess is that a harder pluck is a bigger force, so the disturbance should be driven along faster. That confuses two different speeds. The speed of the wave is how fast the shape slides along the string. The speed of the string is how fast a piece of it moves up and down. A harder pluck raises the second: at 2.5 Hz the angular frequency is $\omega=2\pi\times2.5=15.7$ rad/s, so the peak sideways speed is $A\omega$, which is $0.02\times15.7=0.314$ m/s for the 0.02 m pluck and $0.08\times15.7=1.26$ m/s for the 0.08 m pluck. It does not raise the first, because the shape is not being pushed along the string. The points exchange sideways pulls with their neighbours, and multiplying every displacement by the same number leaves the lag from one point to the next unchanged.

Writing the travelling shape with amplitude $A$ makes the cancellation plain. The derivatives bring down $\omega^2$ and $k^2$, but each still multiplies the same $A\sin(kx-\omega t)$, so $A$ divides out of the wave equation. Here $T$ is the tension and $\mu$ the mass per unit length. The same thing appears in the two-standing-shapes identity: $\sin(kx)\cos(\omega t)+\cos(kx)\sin(\omega t)=\sin(kx+\omega t)$. A harder pluck multiplies the whole right side by $A$, giving $A\sin(kx+\omega t)$. A crest is where $kx+\omega t$ is constant, so differentiating $kx+\omega t=\text{constant}$ gives $k\,dx/dt+\omega=0$, hence the crest speed is $\omega/k$. The amplitude is not in that last equation. It can make the wave taller and make each piece of string move faster sideways, but it cannot make the crest advance faster.

This is exact for an ideal string in the small-deflection limit, where the tension is taken as constant and the slope is small. Pull the middle of a 2 m string to 0.5 m and the string is materially longer than its straight length, so the tension is higher than it was at rest; then $v=\sqrt{T/\mu}$ rises as well. A very hard pluck therefore can change the speed, not because amplitude appears in the linear relation, but because it has changed the tension, which is one of the quantities that sets the speed. Real strings also resist bending, so different wavelengths travel at slightly different speeds, and a disturbance made of several wavelengths changes shape as it goes. The clean statement is the linear one: for small vibrations on a given string, amplitude changes the size and energy of the wave, not its speed.