Answer:
The value is [tex]v_o = 5.488 \ m/s[/tex]
Explanation:
From the question we are told that
The emitted frequency increased by [tex]\Delta f = 1.6 \% = 0.016 \[/tex]
Let assume that the initial value of the emitted frequency is
[tex]f = 100\% = 1[/tex]
Hence new frequency will be [tex]f_n = 1 + 0.016 = 1.016[/tex]
Generally from Doppler shift equation we have that
[tex]f_1 = [\frac{ v \pm v_o}{v \pm + v_s } ] f[/tex]
Here v is the speed of sound with value [tex]v = 343 \ m/s[/tex]
[tex]v_s[/tex] is the velocity of the sound source which is [tex]v_s = 0 \ m/s[/tex] because it started from rest
[tex]v_o[/tex] is the observer velocity So
Generally given that the observer id moving towards the source, the Doppler frequency becomes
[tex]f_1 = [\frac{ v + v_o}{v + 0 } ] f[/tex]
=> [tex]1.016 = [\frac{ 343 + v_o}{343 } ] * 1[/tex]
=> [tex]v_o = 5.488 \ m/s[/tex]
