The speed of a wave is a property of the medium, it is not affected by the length of the string. The frequency may change, but the speed remains constant. There is no change in the speed.

Since the speed of the wave does not change based on the length of the string and we know it has a fundamental frequency of 440Hz, a string of half the length will vibrate at twice the frequency, 880Hz. This makes sense as it will sound higher in pitch. You can try this with a rubber band on a shoebox. Plucking it while placing your finger halfway along the band will result in a higher pitched sound.

The wavelength of the second harmonic of a standing wave on a string is just the length of the string. For the second harmonic, an entire cycle occurs on the length of the string. Therefore, the wavelength of the second harmonic for this string is 37cm or 0.37m. The wavelength for the first harmonic, or fundamental, is twice the length of the string, as this is when one half a cycle occurs over the length of the string.

Harmonics describe the relationship between wavelengths and frequency on a string. A given string will have a wave speed associated with it. Given this wave speed, the harmonics are the frequencies at which half-wavelengths occur along the length of the string. For instance, the first harmonic is the frequency at which a half-wavelength occurs over the string. The second harmonic completes one wave over the string. The string is 1.5 times the wavelength of the third harmonic and so on. Therefore, the  harmonic has a frequency of .

The formula for intensity,  is:

Where  is power in watts and  is the surface area. The surface area  in our case is: 

The power  can be given as  or in our case:

Solve for intensity.

The formula for sound level  is:

, where  is the intensity and  is the threshold of hearing, which is 

First we need to solve for intensity , given by: 

, where  is power,  is the distance from the source of the sound. 

In our case,  and , therefore

To solve for sound level , we do

, where is the intensity and  is the threshold of hearing. 

In this problem, 

, and 

Recall that the formula for sound level  given in decibels is given by:

, where  is the intensity and  is the threshold of hearing. 

Sound level is proportional to intensity by:

If the intensity is increased by a factor of , sound level would increase by a factor of 

In order to find the velocity needed, we use the Doppler equation:

Where  is the speed of sound in air,  is the velocity of the observer, and  is the velocity of the source, which is zero in this case since the guitar player is not moving. Since the guitarist is sharp, your adjustment must decrease the perceived frequency by , which means we use the negative sign in the numerator of the Doppler equation. Plug in the given values into the equation and solve for .

Because we need the note to be the final note to be lower in frequency than the original frequency, we need to run away from it.

The speed of sound is fastest in the least compressible media of the lowest density. Sound does not propagate in a vacuum. Air and water are compressible media, so sound does not travel as fast in these as it does in glass, an incompressible medium. In general, the speed of sound is greatest in solids, and within each phase, faster as density decreases.

We simply need to know the Doppler effect equation to solve this problem:

Since the observer (you) is not moving, we can rewrite the equation as:

The reason for using addition in the denominator is explained below.

We know all of our values, so we can simply solve for the perceived frequency:

For the Doppler effect equation, don't waste your time trying to memorize whether the signs should be . We have purposely made them the same to emphasize the following idea. Think about the situation practically. The police car is moving away from you. Therefore, we know that the percieved frequency is going to be less than the source. How do we make the frequency lower? We either lower the numerator or increase the denominator. Since we can only manipulate the denominator, we will use the + sign. Knowing this will also help you immediately eliminate all answers in which the frequency is increased.