Question

Two cars are approaching each other with same speed of 20 m/s. A man in car A fires bullets at regular intervals of 10 seconds. What will be the time interval noted by a man in car B between 2 bullets?

(velocity of sound = 340 m/s)

• A. 11.1 s
• B. 10 s
• C. 8.9 s
• D. 12 s

Answer 1

Answer: (C)

8.9 s

Answer 2

The problem involves the Doppler effect for sound waves.

Given:

• Speed of car A (source, $v_S$) = 20 m/s
• Speed of car B (observer, $v_O$) = 20 m/s
• Speed of sound ($v$) = 340 m/s
• Time interval between firing bullets (source's time period, $T_S$) = 10 s

We need to find the time interval noted by the man in car B (observed time period, $T_O$).

Since the cars are approaching each other, the observed frequency ($f_O$) will be higher than the source frequency ($f_S$). Consequently, the observed time period ($T_O$) will be shorter than the source time period ($T_S$).

The Doppler effect formula for frequency when the source and observer are approaching each other is:

$f_O = f_S \left( \frac{v + v_O}{v - v_S} \right)$

We know that frequency $f = 1/T$. Substituting this into the formula:

$\frac{1}{T_O} = \frac{1}{T_S} \left( \frac{v + v_O}{v - v_S} \right)$

Rearranging to solve for $T_O$:

$T_O = T_S \left( \frac{v - v_S}{v + v_O} \right)$

Now, substitute the given values:

$T_S = 10$ s $v = 340$ m/s $v_S = 20$ m/s $v_O = 20$ m/s

$T_O = 10 \left( \frac{340 - 20}{340 + 20} \right)$

$T_O = 10 \left( \frac{320}{360} \right)$

$T_O = 10 \left( \frac{32}{36} \right)$

$T_O = 10 \left( \frac{8}{9} \right)$

$T_O = \frac{80}{9}$

Calculating the numerical value:

$T_O \approx 8.888...$ s

Rounding to one decimal place, $T_O \approx 8.9$ s.