We know from the question that

\(\displaystyle \lambda=3.5m\)

\(\displaystyle T=0.01s\)

Frequency is the inverse of period:

\(\displaystyle f=\frac{1}{T}=\frac{1}{0.01s}=100s^{-1}}\)

The velocity of a wave is its wavelength multiplied by its frequency:

\(\displaystyle v=\lambda f = (3.5 m)(100s^{-1})=350\frac{m}{s}\)

The doppler effect follows this formula:

\(\displaystyle f=\frac{s}{s+v}f_0\)

In this equation, \(\displaystyle f\) is the new frequency you will hear, \(\displaystyle s\) is the speed of sound, \(\displaystyle v\) is the velocity of the moving sound-emitting thing, and \(\displaystyle f_0\) is the initial frequency of the sound.

Plugging the given values in, we can describe the initial situation as:

\(\displaystyle 500=\frac{340}{340-30}f_0\)

Note that the velocity is negative \(\displaystyle (-30)\) because the car is driving towards you.

Therefore,

\(\displaystyle f_0=456s^{-1}\)

When the police car is driving away, the situation is described with a positive velocity:

\(\displaystyle f=\frac{340}{340+30}456\)

Therefore,

\(\displaystyle f=419s^{-1}\)

Two waves will emit a beat with a fequency equal to the difference in frequency of the two waves. In this case the beat frequency is:

\(\displaystyle 555 - 550=5 \textup{\textup{ Hz}}\)    (beats per second)

Convert to beats per minute:

\(\displaystyle 5 \textup{\textup{ Hz}}\times \frac{60}{1} \frac{seconds}{minutes} = 300 \textup{\textup{ bpm}}\)

The speed of sound in air is constant, assuming that the temperature of the air is constant. When the length of the string is shortened, by the principles of standing waves, this creates a higher frequencies. Assuming that the two instruments use the same strings is equivalent to stating that the two instruments have strings of equal linear mass density. This situation represents a standing wave, thus we can relate the following equation for the first harmonic:

\(\displaystyle L=2\lambda\)

Where, \(\displaystyle L\) is the length of the string and \(\displaystyle \lambda\) is the wavelength. Then we can use the following equation to relate wavelength and speed (which is known) to frequency:

\(\displaystyle v=\lambda f\)

Since the velocity of sound in a fixed medium is constant, we see that a shorter length, \(\displaystyle L\) corresponds to a shorter wavelength, \(\displaystyle \lambda\). Thus when \(\displaystyle \lambda\) decreases, frequency, \(\displaystyle f\) must increase, to keep velocity constant. 

Intensity is related to radius by the inverse square law: 
\(\displaystyle \frac{I_1}{I_2}=\frac{r_2^2}{r_1^2}\)
This equation is derived from the concept that the energy from the sound waves is conserved and spread out over an area, producing the \(\displaystyle r^2\) term. Applying this concept, when the radius doubles, the intensity decreases by a factor of 4. The correct answer is \(\displaystyle 25W\). 

\(\displaystyle L=0.1725m\)A stopped pipe can be modeled with the following equation: 
\(\displaystyle f_n=\frac{nv}{4L}\)

Rearrange the equation to solve for L, then plug in given values and solve.

\(\displaystyle L=\frac{nv}{4f_n}\)

\(\displaystyle L=\frac{1*345}{4*500}\)

\(\displaystyle L=0.1725m=17.25cm\)

Frequency of beats is determined by the absolute value of the difference between two different frequencies. Thus, the beats frequency is 2Hz. Note that beat frequency is always a positive number.

An open pipe can be modeled by the following equation: 

\(\displaystyle f_n=\frac{nv}{2L}\)

Rearrange the equation to solve for \(\displaystyle L\) then plug in given values and solve.

\(\displaystyle L=\frac{nv}{2f_n}\)

\(\displaystyle L=\frac{1*345}{2*600}\)

\(\displaystyle L=0.2875m=28.25cm\)

Sound is a longitudinal, or compression wave. A region of slightly more compressed air is followed by a region of slightly less compressed air (called a rarefaction). When the compressed air is behind the balloon, it pushes it forward, and when it is in front of the balloon, it pushes it back. This only works if the frequency is low, because the waves are long enough so that the balloon can react to them.

Use the following equation to find the velocity of the wave, using its fundamental frequency and the length:

\(\displaystyle f_1=\frac{v}{2L}\)

\(\displaystyle v=\frac{440}{2*0.37}=325.6\frac{m}{s}\)