# In one of the original Doppler experiments, one tuba was played on a moving platform car at a...

## Question:

In one of the original Doppler experiments, one tuba was played on a moving platform car at a frequency of {eq}75 \ Hz {/eq}, and a second identical one was played on the same tone while at rest in the railway station. What beat frequency was heard if the train approached the station at a speed of {eq}18.0 \ m / s {/eq}?

## Beat Frequency from Doppler Shift:

Just like any other wave, sound waves also produce constructive and destructive interference. Constructive interference creates an intensity maximum and destructive interference creates an intensity minimum. A maximum and minima of sound happen alternatively and a pair of maxima and minima are treated as one beat. The number of beats formed is equal to the difference in frequency between the interfering sound waves. When the source of sound moves towards the listener with a speed {eq}v_1 {/eq}, the observed frequency of sound {eq}F_1 =\dfrac { v } { v - v_1 } F {/eq}. Here {eq}v, \ \ F {/eq} are the speed of sound in air and original frequency produced by the source respectively.

Given data

• Original frequency played by the tuba {eq}F = 75 \ Hz {/eq}
• The speed with which the train carrying the tube approaches the station {eq}v_1 = 18 \ m/s {/eq}
• Speed of sound in air {eq}v = 343 \ m/s {/eq}

Observed frequency of the approaching tuba in the train {eq}F_1 = \dfrac { v } { v - v_1 } F \\ F_1 = \dfrac { 343 } { 343 - 18 } \times 75 \\ F_1 = 79.15 \ Hz {/eq}

Beat frequency produced between the two tuba {eq}F_b = F_1 - F \\ F_b = 79.15 - 75 \\ F_b = 4.15 \ Hz {/eq} 