Question

How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 1.6 % increase in the emitted frequency?

Answer 1

The value is
`v_o = 5.488 \ m/s`

Step-by-step explanation:

From the question we are told that

The emitted frequency increased by
`\Delta f = 1.6 \% = 0.016 \`

Let assume that the initial value of the emitted frequency is

`f = 100\% = 1`

Hence new frequency will be
`f_n = 1 + 0.016 = 1.016`

Generally from Doppler shift equation we have that

`f_1 = [( v \pm v_o)/(v \pm + v_s ) ] f`

Here v is the speed of sound with value
`v = 343 \ m/s`

`v_s` is the velocity of the sound source which is
`v_s = 0 \ m/s` because it started from rest

`v_o` is the observer velocity So

Generally given that the observer id moving towards the source, the Doppler frequency becomes

`f_1 = [( v + v_o)/(v + 0 ) ] f`

=>
`1.016 = [( 343 + v_o)/(343 ) ] * 1`

=>
`v_o = 5.488 \ m/s`