# A physics professor demonstrates the Doppler effect by tying a 550 Hz sound generator to a 1 m long rope and whirling it around her head in ahorizontal circle at 120 rpm. Assume the room...

A physics professor demonstrates the Doppler effect by tying a 550 Hz sound generator to a 1 m long rope and whirling it around her head in a

horizontal circle at 120 rpm. Assume the room temperature is 20 *C. A) What is the highest frequency heard by a student in the classroom? B) What is the lowest frequency heard by a student in the classroom?

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The formula for Doppler Effect is:

`f ' = v(f/(v+-u_s))`

where f ' - the apparent frequency (heard by the observer)

f - frequency of source

`u_s` - speed of source

v - speed of sound

Since the sound source is moving in a circular motion with an angular speed of 120 rpm, the linear speed must be determined.

`u_s = omega *r = [120 (rev)/(min) * (1 min)/(60 sec) * (2pi rad)/(1rev)] * 1m `

`u_s = 12.6 m/s`

Hence, `u_s = 12.6 m/s`

The speed of sound at a temperature `20^oC` must be determined, too. The formula for speed of sound in air at a given temperature is:

`v = 331.4 + 0.6T_c `

where `T_c` -temperature in Celsius.

`v = 331.4 + 0.6(20) = 331.4 + 12 = 343.4 m/s`

So, at `20^oC` `v= 343.4 ` m/s.

Then, substitute values of v, `u_s ` and f to the formula for Doppler Effect to solve for the highest and lowest apparent frequency.

(A) Apparent freuqency is highest when the sound source is moving toward the observer. To solve, sign of `u_s` is negative.

`f ' = v(f/(v-u_s)) = 343.4 (550/(343.4-12.6))`

`f ' = (343.4*550)/330.8= 188870/330.8 = 570.9 Hz`

**Thus ,highest frequency heard by the student is 570.9 Hz.**

(B) Apparent frequency is lowest when the sound source is moving away from the observer. In this case, sign of `u_s` is positive.

`f ' = v(f/(v+u_s)) = 343.4(550/(343.4+12.6))`

`f ' = (343.4*550)/355.4 = 188870= 531.4Hz.`

**Hence, lowest frequency heard by the student is 531.4 Hz.**

**Sources:**