Question

Two tuning forks with natural frequencies 340 H_(Z) each move relative to a stationary observer . One forks moves away from the oberver while the other moves towards him at the same speed . The observer hearts beats of frequency 3 H_(Z) . Find the speed the of the tuning fork (velocity of sound in air is 340 m//s) .

Answer 1

Speed of the tuning fork:

Given,  $f_1 - f_2 = 3$

⇒ $(\frac{v}{v-v_s} )f-(\frac{v}{v+v_s}) f=3$

⇒$\frac{340*340}{340-v_s} - \frac{340*340}{340+v_s} = 3$

⇒$340( \frac{1}{1-\frac{v_s}{340} } - \frac{1}{1+\frac{v_s}{340} })=3$

⇒$340((1-\frac{v_s}{340} )^{-1} - (1+\frac{v_s}{340} )^{-1})=3$

Using Binomial Expansion and neglecting higher orders, we get:

   $340((1+\frac{v_s}{340} ) - (1-\frac{v_s}{340} ))=3$

⇒$340*2v_s = 3*340$

⇒ $v_s =1.5 m/s$

Answer 2

Here is your answer:

Explanation:

Given,

f1−f2=3

or (ν/ν−νs)f−(ν/ν+νs)f=3

or [1(ν−νs/ν)−1(1+νs/ν)]f = 3

or [(1−νs/ν)−1−(1+νs/ν)−1]f = 3

or [(1+νs/ν)−(1−νs/ν)]f = 3

or 2νsf/ν = 3

or Speed of tuning fork, νs = 3ν/2f

substituting the values, we get

νS=(3)(340)/(2)(340)

    = 1.5m/S