Wave on a string to sound wave If you have a string of 2m in length, and the wave speed on the string is 2m/s. and when then string vibrates at fundamental frequency the wavelength of the wave would then be 4m.
However, the sound created by the vibrating string would have a speed of around 340m/s (speed of sound depending on temperature). Since the velocity increased this means either the wavelength or the frequency must change because:
$$v = \lambda f$$
My question is, would the wavelength increase, or the frequency, or both? If you can explain a bit that would be even better.
 A: Any confusion arises because the speed of sound on the string is different in the air than it is on the string.
It is true that if you have a string of 2m in length fixed at the ends, and the wave speed on the string is 2m/s, then the wavelength on the string is 4m (which means that there is a one half wavelength standing wave on the string), and the frequency of the vibration is 2m/s / 4m = 0.5 Hz. That is a very low frequency, too low to hear.  Nevertheless, the string would push on the air as it vibrates setting up a sound wave in the air, also at 0.5 Hz.  That wave would travel through the air at the speed of sound in the air, which is about 350 m/s, and it would have a very long wavelength in the air of about 350/.5= 700m.  But it is too low a frequency for us to hear.
To take a more realistic example with a more realistic speed of sound on a string, consider the lowest pitched string on a violin.  It has a length of 0.33m and it is designed to have a mass and tension so that the speed of the wave along the string is about 129 m/s. The wavelength is twice the string length, or 0.66 m. It then vibrates with a pitch of the G below middle C, or 196Hz (that is, 129/.66).  As it vibrates, it pushes the air around, and this time the pitch is great enough for us to easily hear.  The speed of sound in the air is 350 m/s, the vibration frequency is 196Hz, so the wavelength in the air is about 1.76m.
A: The frequency of a wave cannot easily change: to do so, you would have to somehow "store the phase difference" between points. What I mean is: if one point is generating 10 wave cycles per second, and another point observes only 5, there must be five wave cycles "not yet observed". This is exactly what happens in the Doppler effect: if you move away from a sound source, there are "waves on their way" that haven't reached you yet, which is why you can observe a different frequency than the one emitted.
But back to your string: if both source and observer are stationary, they must observe the same frequency. And so, if the velocity of the sound wave is different in the string and in the air, then the wave length in the air will also be different.
