JEE PYQ: Waves And Sound Question 15
Question 15 - 2020 (05 Sep Shift 2)
Two coherent sources of sound, $S_1$ and $S_2$, produce sound waves of the same wavelength, $\lambda = 1$ m, in phase. $S_1$ and $S_2$ are placed 1.5 m apart (see fig.). A listener, located at $L$, directly in front of $S_2$ finds that the intensity is at a minimum when he is 2 m away from $S_2$. The listener moves away from $S_1$, keeping his distance from $S_2$ fixed. The adjacent maximum of intensity is observed when the listener is at a distance $d$ from $S_1$. Then, $d$ is :
(1) 12 m
(2) 5 m
(3) 2 m
(4) 3 m
Show Answer
Answer: (4)
Solution
Initially, $S_2 L = 2$ m
$S_1 L = \sqrt{2^2 + \left(\frac{3}{2}\right)^2} = \frac{5}{2} = 2.5$ m
Path difference, $\Delta x = S_1 L - S_2 L = 0.5$ m $= \frac{\lambda}{2}$
When the listener moves from $L$, first maxima will appear if path difference is integral multiple of wavelength.
$\Delta x = n\lambda = 1\lambda$ ($n = 1$ for first maxima)
$\therefore \Delta x = \lambda = S_1 L’ - S_2 L$
$\Rightarrow 1 = d - 2 \Rightarrow d = 3$ m
BETA
