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

Answer:

The value is  $v_o = 5.488 \ m/s$

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

         $f_1 = [\frac{ v \pm v_o}{v \pm + v_s } ] f$

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

                   $f_1 = [\frac{ v + v_o}{v + 0 } ] f$

=>                $1.016 = [\frac{ 343 + v_o}{343 } ] * 1$  

=>                $v_o = 5.488 \ m/s$