Question 169: Waves

Question

A stationary observer receives sound from two identical tuning forks, one of which approaches
and the other one recedes with the same speed (much less than the speed of sound). The
observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0
 = 1400 Hz and the
velocity of sound in air is 350 m/s. The speed of each tuning fork is close to :

Accepted Answer

f
1
= $$\left( {{c \over {c - v}}} ight){f_0}$$
f
1
= $$\left( {{c \over {c + v}}} ight){f_0}$$
beat frequency = f
1
Ã¢ÂÂ f
2
= $$c{f_0}\left( {{1 \over {c - v}} - {1 \over {c + v}}} ight)$$
= $$c{f_0}\left( {{{2v} \over {{c^2} - {v^2}}}} ight)$$
As c $$ \gg $$ v then $${{c^2} - {v^2}}$$ = $${{c^2}}$$
= $$c{f_0}\left( {{{2v} \over {{c^2}}}} ight)$$
= $${f_0}\left( {{{2v} \over c}} ight)$$
$$ 	herefore $$ $${f_0}\left( {{{2v} \over c}} ight)$$ = 2
$$ \Rightarrow $$ v = $${{350} \over {1400}}$$ = $${1 \over 4}$$ m/s

Answer

1 m/s

Answer

$${1 \over 8}$$ m/s

Answer

$${1 \over 4}$$ m/s

Answer

$${1 \over 2}$$ m/s

Waves

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