MCQ
Two sources $A$ and $B$ are sounding notes of frequency $660 \,Hz$. $A$ listener moves from $A$ to $B$ with a constant velocity $u$. If the speed of sound is $330 \,m / s$, what must be the value of $u$ so that he hears $8$ beats per second is .......... $m / s$
  • A
    $2.8$
  • $2$
  • C
    $3.0$
  • D
    $3.5$

Answer

Correct option: B.
$2$
b
(b)

Apparent frequency of sound heard by the observer from source is

$n_1=\left(\frac{v-v_0}{v-v_s}\right) n$

$=\left(\frac{v-u}{v+v_s}\right) n$

Apparent frequent of sound heard by the observe from source

$n _2=\left(\frac{ v + v _0}{ v - v _0}\right) n$

$=\left(\frac{ v + u }{ v - v _{ s }}\right) n$

$\text { No. of beats }=8$

$n _2- n _{ l }=8$

$\left(\frac{ v + u }{ v - v _{ s }}\right) n -\left(\frac{ v - u }{ v + v _{ s }}\right) n =8 \quad v =300, v _{ s }=0, n =660$

$\Rightarrow\left(\frac{330+ u }{300-0}\right)(600)-\left(\frac{300- u }{330+0}\right) 660=8$

$\therefore \frac{2 \times 6604}{330}=8$

$4 u =8$

$u =2 .$

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