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Two small stereo speakers A and B that are 1.40 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point P starts out equidistant from both speakers and walks so that he is always 1.50 m from speaker B (Fig. E35.1). For what values of x will the sound this person hears be (a) maximally reinforced, (b) cancelled? Limit your solution to the cases where x ≤1.50 m.

Short Answer

Expert verified

(a) The values of x will the sound this person hears be maximally reinforced is 150cm,116cm,82cm,48cm,14cm.

(b) The values of x will the sound this person hears is cancelled is 133cm,99cm,65cm,31cm.

Step by step solution

01

(a) Determination of the values of x will the sound this person hears be maximally reinforced.

The path length the person took denoted by x is in the range of 0 ≤x≤150 cm.

The sound is reinforced to its maximum when the path difference between the sound waves is an integral multiple of the wavelength. This is also known as the Constructive interference. The condition is,

r2-r1=³¾Î»,      m=0,±1,±2,... ...(i)

Here, in this problem r2= 150 cm and r2= x cm.

Substitute the values in equation (i),

150cm-x=m(34cm)x=150cm-m(34cm)

Thus, the values of x are,

localid="1663912422978" form=0x=150cm

form=1x=150cm-34cm=116cm

localid="1663912531079" form=2x=150cm-2(34)cm=82cm

form=3x=150cm-3(34)cm=48cm

form=4x=150cm-4(34)cm=14cm

02

(b) Determination of the values of x will the sound this person hears is cancelled.

The sound is cancelled when the path difference between the sound waves is a multiple of the odd number of half wavelengths. This is also known as the destructive interference. The condition is,

r2-r1=m+12λ,      m=0,±1,±2,... ...(ii)

Here, in this problemr2= 150 cm andr1= x cm.

Substitute the values in equation (ii),

150cm-x=m+12(34cm)x=150cm-m+12(34 cm)

form=0x=133cmform=1x=150cm-3234cm=99cmform=2x=150cm-5234cm=65cmform=3x=150cm-7234cm=31cm

Thus, The values of x for which the sound will be cancelled is 133 cm, 99 cm, 65 cm, 31 cm.

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