Question

Mike is playing his speaker at an unknown frequency. His friend, Emma, gets on her bike and rides away from Mike at a uniform speed, holding an open-air column horizontally over her head. What was the frequency of the sound coming from Mike's speaker?

Mike is playing his speaker at an unknown frequency on a hot `31.0^@C` day. His friend, Emma, gets on her electric bike and rides away from Mike with a uniform speed of `16.0m/s`, holding a `0.450m` open-air column horizontally over her head. The air rushing through the tube produces a third harmonic tone in the tube. Mike hears a beat frequency of `5.00Hz`. What was the frequency of the sound coming out of Mike's speaker?

Answer 1

`1109.33Hz` or `1119.33Hz`
(significant figures excluded)

Explanation:

Order of Solving

`1`. Determine the speed of the sound waves coming from Mike's speaker according to an air temperature of `31.0^@C`.

`2`. Find the frequency produced from the sound waves going through the tube above Emma's head according to a third harmonic tone.

`3`. Find the frequency Mike detects when the sound waves travel through to him as Emma rides away.

`4`. Determine the possible frequencies that could have come out of Mike's speaker.

Step 1
The speed of sound waves in air depends on the air temperature. The higher the temperature, the greater the speed. At `0^@C`, the speed of sound is `331m/s`. Each increase in one degree Celsius results in a speed increment of `0.6m/s`. This can be represented by the formula:

`color(blue)(|bar(ul(color(white)(a/a)color(black)(v_s=331m/s+((0.6m/s)/(color(white)(i)^@C))xx"temperature")color(white)(a/a)|)))`

Using the formula, the speed of the sound waves coming from Mike's speaker was

`color(darkorange)(v_s)=331m/s+((0.6m/s)/(color(white)(i)^@C))xx31.0^@C`

`=color(darkorange)(349.6m/s)`

Step 2
The air travelling from Mike's speaker through the tube above Emma's head produces a third harmonic tone. Since the information regarding the tube's length and harmonic is given, we can use that knowledge to combine it with the speed of the sound waves to determine the frequency of the sound coming from the tube. This can be found by using the formula for the frequency of standing waves in an air column open at both ends:

`color(blue)(|bar(ul(color(white)(a/a)color(black)(f=(nv)/(2L))color(white)(a/a)|)))`

`ul("where:")`
`f=`frequency
`n=`harmonic number
`v=`speed of wave
`L=`length of air column

In our case, we will let `f` be `f_t`, the frequency of the tube, and `v` be `v_s`, the speed of the sound waves.

`color(purple)(f_t)=(ncolor(darkorange)(v_s))/(2L)`

`=((3)(color(darkorange)(349.6m/s)))/(2(0.450m))`

`=color(purple)(1165.33Hz)`

Step 3
The Doppler effect causes a change in the frequency heard by Mike as Emma rides away from him. The frequency that he hears is lower than the actual frequency of the sound waves in the tube. To derive the frequency detected by Mike, we can use the Doppler effect formula for sound:

`color(blue)(|bar(ul(color(white)(a/a)color(black)(f_d=(v_s/(v_s+-v_o))f)color(white)(a/a)|)))`

`ul("where:")`
`f_d=`detected frequency
`v_s=`speed of wave
`v_o=`speed of object
`f=`actual frequency

In our case, we will let `f` be `f_t`. The sign between `v_s` and `v_o` will be `+` and not `-` since the tube is moving away from Mike.

`color(teal)(f_d)=(v_s/(v_s+v_o))color(purple)(f_t)`

`=((349.6m/s)/(349.6m/s+16m/s))color(purple)(1165.33Hz)`

`=color(teal)(1114.33Hz)`

Step 4
The frequency detected by Mike is different from the actual frequency of the tube, but it's also different from the frequency coming out of his speaker. Mike reports that he hears a beat frequency of `5.00Hz`, so we can use the beat frequency formula to determine the two possible frequencies coming out of his speaker:

`color(blue)(|bar(ul(color(white)(a/a)color(black)(f_b=|f_1-f_2|)color(white)(a/a)|)))`

`ul("where:")`
`f_b=`beat frequency
`f_1=`first frequency
`f_2=`second frequency

In our case, we will let `f_2` be `f_d`, the frequency detected by Mike. The unknown variable, `f_1`, is the frequency of the sound coming from the speaker.

Since the equation has an absolute value sign, there are two possible scenarios.

`ul("Case 1:")`
`f_b=f_1-color(teal)(f_d)`

`5.00Hz=f_1-color(teal)(1114.33Hz)`

`f_1=color(green)(|bar(ul(color(white)(a/a)color(black)(1119.33Hz)color(white)(a/a)|)))`

`ul("Case 2:")`
`f_b=-(f_1-color(teal)(f_d))`

`5.00Hz=-(f_1-color(teal)(1114.33Hz))`

`f_1=color(green)(|bar(ul(color(white)(a/a)color(black)(1109.33Hz)color(white)(a/a)|)))`