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belka [17]
2 years ago
9

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?
Physics
1 answer:
morpeh [17]2 years ago
8 0

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

         

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