JEE PYQ: Waves And Sound Question 19
Question 19 - 2020 (07 Jan Shift 2)
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 $v_0 = 1400$ Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to:
(1) $\frac{1}{2}$ m/s
(2) 1 m/s
(3) $\frac{1}{4}$ m/s
(4) $\frac{1}{8}$ m/s
Show Answer
Answer: (3)
Solution
From Doppler’s effect, frequency of sound heard ($f_1$) when source is approaching:
$f_1 = f_0 \frac{c}{c - v}$
Frequency of sound heard ($f_2$) when source is receding:
$f_2 = f_0 \frac{c}{c + v}$
Beat frequency $= f_1 - f_2$
$\Rightarrow 2 = f_1 - f_2 = f_0 c\left[\frac{1}{c - v} - \frac{1}{c + v}\right]$
$= f_0 c \cdot \frac{2v}{c^2 - v^2}$
For $c \gg v$:
$v = \frac{2c}{2f_0} = \frac{c}{f_0} = \frac{350}{1400} = \frac{1}{4}$ m/s
BETA
