The frequency of a tuning fork is 256 Hz. The velocity of sound in air 344 text ms-1. The distance travelled (in metres) by the sound during the time in which the tuning fork completes 32 vibrations is:

Question

The frequency of a tuning fork is 256 Hz. The velocity of sound in air 344 text ms-1. The distance travelled (in metres) by the sound during the time in which the tuning fork completes 32 vibrations is: (A) 21 (B) 43 (C) 86 (D) 129

Tardigrade Admin · Accepted Answer

Wavelength of sound, λ =(v/n)=(344/256)m So, distance travelled by the sound in 32 vibrations is s=32λ =32× (344/256)=43 m

Q. The frequency of a tuning fork is 256 Hz. The velocity of sound in air $344 m s^{- 1} .$ The distance travelled (in metres) by the sound during the time in which the tuning fork completes 32 vibrations is:

21

43

86

129

Wavelength of sound, $λ = \frac{v}{n} = \frac{344}{256} m$ So, distance travelled by the sound in 32 vibrations is $s = 32 λ$ $= 32 \times \frac{344}{256} = 43 m$

Q. The frequency of a tuning fork is 256 Hz. The velocity of sound in air 344ms−1. The distance travelled (in metres) by the sound during the time in which the tuning fork completes 32 vibrations is:

21

43

86

129

Wavelength of sound, λ=nv​=256344​m So, distance travelled by the sound in 32 vibrations is s=32λ =32×256344​=43m

Q. The frequency of a tuning fork is 256 Hz. The velocity of sound in air $344 m s^{- 1} .$ The distance travelled (in metres) by the sound during the time in which the tuning fork completes 32 vibrations is:

Wavelength of sound, $λ = \frac{v}{n} = \frac{344}{256} m$ So, distance travelled by the sound in 32 vibrations is $s = 32 λ$ $= 32 \times \frac{344}{256} = 43 m$