The wavelengths of two notes in the air are (36/195) m and (36/193) m . Each note produces 10 beats per second separately with a third note of fixed frequency. The velocity of sound in air in m s- 1 is

Question

The wavelengths of two notes in the air are (36/195) m and (36/193) m . Each note produces 10 beats per second separately with a third note of fixed frequency. The velocity of sound in air in m s- 1 is (A) 330 (B) 340 (C) 350 (D) 360

Tardigrade Admin · Accepted Answer

Beat frequency= v1∼ v2 Let the frequency of third note be n . Then, (195 v/36)-n=10 ldots (i) And n-(193 v/36)=10 ldots (i i) Adding Eqs. (i) and (ii) (v/18)=20 Longrightarrow v=360 ms- 1

Q. The wavelengths of two notes in the air are $\frac{36}{195} m$ and $\frac{36}{193} m$ . Each note produces $10$ beats per second separately with a third note of fixed frequency. The velocity of sound in air in $m s^{- 1}$ is

$330$

$340$

$350$

$360$

Beat frequency= $v_{1} \sim v_{2}$
Let the frequency of third note be $n$ .
Then,
$\frac{195 v}{36} - n = 10 \dots (i)$
And
$n - \frac{193 v}{36} = 10 \dots (ii)$
Adding Eqs. (i) and (ii)
$\frac{v}{18} = 20$
$⟹ v = 360 m s^{- 1}$

Q. The wavelengths of two notes in the air are 19536​m and 19336​m . Each note produces 10 beats per second separately with a third note of fixed frequency. The velocity of sound in air in ms−1 is

330

340

350

360