Question

8. The apparent frequency noted by a moving listener
away from the stationary source is 10% less than the
real frequency. If the velocity of sound is 330 m/s,
then the velocity of the listener is
a) 8.5 m/s (b) 40 m/s (c) 50 m/s
53 m
d) 33 m/s​

Answer 1

Given:

Apparent frequency (f) noted by a moving listener away from the stationary source = 10% less than the real frequency (f₀)

Velocity of sound (v) = 330 m/s

Velocity of source $\sf (v_s)$ = 0 m/s (Stationary)

To Find:

Velocity of listener (v₀)

Answer:

According to the question;

f = 90% f₀

f = 0.9 f₀

Doppler's Effect:

$\boxed{ \boxed{ \bf{f = f_0 \bigg( \dfrac{v \pm v_0}{v \pm v_s} \bigg)}}}$

As the listener is moving away and source is stationary. So,

$\bf f = f_0 \bigg( \dfrac{v - v_0}{v } \bigg)$

By substituting values we get:

$\rm \implies 0.9\cancel{f_0} = \cancel{f_0} \bigg( \dfrac{330 - v_0}{330 } \bigg) \\ \\ \rm \implies 0.9 = \dfrac{330 - v_0}{330} \\ \\ \rm \implies 330 - v_0 = 330 \times 0.9 \\ \\ \rm \implies 330 - v_0 = 297 \\ \\ \rm \implies v_0 = 330 - 297 \\ \\ \rm \implies v_0 = 33 \: m {s}^{ - 1}$

$\therefore$ Velocity of listener (v₀) = 33 m/s

Correct Option: $\boxed{\mathfrak{(d) \ 33 \ m/s}}$

Answer 2

90/100 f = f(330-v/330)

v = 330 - 0.9 × 330

v = 33 m/s