Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Two speakers are separated by a distance of $5 \mathrm{~m}$ . A person stands at a distance of $12 \mathrm{~m}$ , directly in front of one of the speakers If $n$ stands for any integer, find the frequencies for Which the listener will hear a minimum sound intensity. speed of sound in air is $340 \mathrm{~m} / \mathrm{s}$
Option: 1
$f=170(2 n+1) Hz$

Option: 2
$f=160(2 n+1) Hz$

Option: 3
$f=160(n+1) Hz$

Option: 4
$f=10(2 n+1) Hz$

Answers (1)

best_answer

$\left(n+\frac{1}{2}\right) \lambda=\text { Path difference } \\$

$\left(n+\frac{1}{2}\right) \lambda=(13-12) \\$

$\left(n+\frac{1}{2}\right) \lambda=1 \\$

$(2 n+1) \lambda=2 \\$

$\lambda=\frac{v}{f}=\frac{340}{f} \\$

$(2 n+1) \times \frac{340}{f}=2 \\$

$f=170(2 n+1) Hz$

Posted by

Rishabh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30

Latest Question