A bat flies perpendicular to a wall at 6m-s- emitting a sound of frequency 4-5- 104 Hz- The number of beats it hears per second is velocity of sound in air - 340 m-s

The correct option is A 1616 Given : v=340 m/s v_s=6 m/s δ=4.5× 10^4 H_z v_0= 6 m/s By doppler effect formula δ^'=δ ( v v_0v v_3 ) ...

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Standard XII

Physics

Intensity of Sound Waves

A bat flies p...

Question

A bat flies perpendicular to a wall at $6 m / s,$ emitting a sound of frequency $4.5 \times 10^{4} H z$ . The number of beats it hears per second is (velocity of sound in air $= 340 m / s$ )

21.37

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1200

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1616

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92.7

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Solution

The correct option is A $1616$
Given : $v = 340 m / s$

$v_{s} = 6 m / s$

$δ = 4.5 \times 10^{4} H_{z}$

$v_{0} = - 6 m / s$

By doppler effect formula

$δ^{^{'}} = δ (\frac{v - v_{0}}{v - v_{3}})$

$= 4.5 \times 10^{4} (\frac{340 - (- 6)}{340 - 6})$

$= 4.5 \times 10^{4} (\frac{346}{334})$

$= 46616$

Beats $= δ^{^{'}} - δ$

$= 46616 - 45000$

$= 1616$

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A bat flies perpendicular to a wall at 6m/s, emitting a sound of frequency 4.5×104 Hz. The number of beats it hears per second is (velocity of sound in air =340m/s)

The correct option is A 1616Given : v=340 m/svs=6 m/sδ=4.5×104 Hzv0=−6 m/sBy doppler effect formulaδ′=δ(v−v0v−v3)=4.5×104(340−(−6)340−6)=4.5×104(346334)=46616Beats =δ′−δ=46616−45000=1616

A bat flies perpendicular to a wall at $6 m / s,$ emitting a sound of frequency $4.5 \times 10^{4} H z$ . The number of beats it hears per second is (velocity of sound in air $= 340 m / s$ )