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Two loudspeakers are placed 3.00 m apart. They emit 435 Hz sounds, in phase. A microphone is...

Question:

Two loudspeakers are placed 3.00 m apart. They emit 435 Hz sounds, in phase. A microphone is placed d = 3.40 m distant from a point midway between the two speakers, where an intensity maximum is recorded.

(a) How far must the microphone be moved to the right to find the first intensity minimum?

(b) Suppose the speakers are reconnected so that the 435 Hz sounds they emit are exactly out of phase. At what positions (how far must the microphone be moved to the right) are the intensity maximum and minimum now?

Intensity Of A Speaker:

The distance until which a given frequency of an object is properly audible for a certain wavelength. The intensity of an object depends upon the wavelength and the frequency of an object. The wavelength of the object is the ratio of the velocity of the sound to the frequency of the given object.The velocity of the sound is 343 m/s.

Answer and Explanation:

The wavelength will be {eq}\lambda=\frac{v}{f}\\ \lambda=\frac{343}{435}\\ \lambda=0.788 m {/eq}

Where v is the velocity of the sound.

The distance will be

{eq}d_1=\sqrt((L/2-x)^2+d^2)\\ d_2=\sqrt((L/2+x)^2+d^2)\\ {/eq}

We know

{eq}d_2-d_1=\frac{\lambda}{2}\\ \sqrt((L/2+x)^2+d^2)-\sqrt((L/2-x)^2+d^2)=\frac{\lambda}{2} {/eq}

Squaring and finding x

{eq}x^2=\frac{\lambda^2L^2/4+\lambda^2d^2-\lambda^4/16}{4L^2-\lambda^2}\\ x^2=\frac{0.788^2\times 3^2/4+0.788^2\times 3.4^2-0.788^4/16}{4(3)^2-0.788^2}\\ x^2=\frac{1.39+7.17-0.024}{29.8}\\ x=0.535 m {/eq}

Therefore the microphone must be moved 0.535 m.


b) Minimum distance {eq}x_{min}=0.535 m {/eq}

We know for maximum

{eq}x_{max}=(n-\frac{1}{2})\frac{\lambda d}{a}\\ n=1\\ x_{max}=\frac{\lambda d}{2a}\\ x_{max}=\frac{0.788\times 3.40}{2\times 3}\\ x_{max}=0.446 m {/eq}


Learn more about this topic:

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Intensity in Physics: Definition & Measurement

from High School Physics: Tutoring Solution

Chapter 5 / Lesson 10
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