(a) Explanation Given info: The two identical speakers are kept at 2.0 m apart, the frequency of the sound produces by the speaker is 784 Hz , the speed of the sound in air is 344 m/s . The following figure shows the arrangement of the speakers and the microphone. Figure-(I) Formula to calculate the path difference for the destructive interface condition is, Δ l = ( m + 1 2 ) ( v f ) (I) Δ l is the path difference. m is an integer and its value varies as m = 0 , 1 , 2 , 3 , 4 , ⋅ ⋅ ⋅ . v is the speed of the sound in air. f is the frequency of the sound. Formula to calculate the path difference from the figure (I) is, Δ l = A P − B P A P is the distance between the point A and point P . B P is the distance between the point B and point P . Substitute ( A B ) 2 + x 2 for A P and x for B P in the above equation to find Δ l . Δ l = ( ( A B ) 2 + x 2 ) − ( x ) Substitute ( ( A B ) 2 + x 2 ) − ( x ) for Δ l in the equation (I) to find x . ( ( A B ) 2 + x 2 ) − ( x ) = ( m + 1 2 ) ( v f ) ( ( A B ) 2 + x 2 ) = ( x ) + ( m + 1 2 ) ( v f ) ( A B ) 2 + x 2 = ( x 2 ) + 2 ( m + 1 2 ) ( v f ) x + ( m + 1 2 ) 2 ( v f ) 2 x = ( A B ) 2 2 ( m + 1 2 ) ( v f ) − ( m + 1 2 ) 2 ( v f ) 2 2 ( m + 1 2 ) ( v f ) (II) For m = 0 , the distance of destructive interface is, Substitute 0 for m , 2.0 m for A B and 344 m/s for v and 784 Hz for f in the equation (II) to find x . x = ( 2.0 m ) 2 2 ( 0 + 1 2 ) ( 344 m/s 784 Hz ) − ( 0 + 1 2 ) 2 ( 344 m/s 784 Hz ) 2 2 ( 0 + 1 2 ) ( 344 m/s 784 Hz ) = 4.0 m 2 0.438 m − 0.047961 m 2 0.438 m = 9.01 m Thus, the distance of destructive interface for m = 0 is 9.01 m . For m = 1 , the distance of destructive interface is, Substitute 1 for m , 2.0 m for A B and 344 m/s for v and 784 Hz for f in the equation (II) to find x . x = ( 2.0 m ) 2 2 ( 1 + 1 2 ) ( 344 m/s 784 Hz ) − ( 1 + 1 2 ) 2 ( 344 m/s 784 Hz ) 2 2 ( 1 + 1 2 ) ( 344 m/s 784 Hz ) = 4 (b) To determine The distances at which constructive interface of the sound occurs. (c) To determine The low frequency of the sound.

University Physics with Modern Physics, Volume 1 (Chs. 1-20) and Mastering Physics with Pearson eText & ValuePack Access Card (14th Edition)

14th Edition

ISBN: 9780134209586

Author: Hugh D. Young

Publisher: PEARSON

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Interference of sound

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Chapter 16, Problem 16.65P

(a)

To determine

The distances at which destructive interface of the sound occurs.

(b)

To determine

The distances at which constructive interface of the sound occurs.

(c)

To determine

The low frequency of the sound.

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University Physics with Modern Physics, Volume 1 (Chs. 1-20) and Mastering Physics with Pearson eText & ValuePack Access Card (14th Edition)

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The distances at which destructive interface of the sound occurs.

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The distances at which destructive interface of the sound occurs.

The distances at which constructive interface of the sound occurs.

The low frequency of the sound.

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