Two point sources of sound S1 and S2 that emit sound of wavelength ? = 2.00 m. The emissions are the same in all directions and in phase, an

Question

Two point sources of sound S1 and S2 that emit sound of wavelength ? = 2.00 m. The emissions are the same in all directions and in phase, and the separation between the sources is d = 10.0 m. At any point P on the x axis, the wave from S1 and the wave from S2 interfere

(a) When P is very far away (x ? ?), what is the phase difference between the arriving waves from S1 and S2? Is the interference they produce constructive, destructive, or intermediate?

Now move point P along the x axis toward S1.

(a) Does the phase difference between the waves increase or decrease?

(b) At what distance x do the waves have a phase difference of 0.50??

(c) At what distance x do the waves have a phase difference of 1.00??

(d) At what distance x do the waves have a phase difference of 1.50??

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Thiên Ân 4 years 2021-08-26T09:17:04+00:00 1 Answers 12 views 0

Answers ( )

  1. RuslanHeatt
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    2021-08-26T09:18:43+00:00
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    Explanation:

    Given , Emitted wavelength , \lambda = 2.00 m

    Distance between sources , d = 10.0 m

    (a) When P is very far , x\rightarrow \infty

    Let , The distance from S1 to the point P = d1 = x

    The distance from S2 to the point P = d2

    The distance d2 can be calculated using pythagoras theorem :

    d_{2}^{2}=d^{2}+d_{1}^{2}

    d_{2}=\sqrt{d^{2}+d_{1}^{2}}

    d_{2}=\sqrt{(10)^{2}+(x)^{2}}

    The difference in distance can be calculated as :

    d_{2}-d_{1}=\sqrt{(10)^{2}+(x)^{2}}-d_{1}

    d_{2}-d_{1}=\sqrt{(10)^{2}+(x)^{2}}-x …………………………… equation (1)

    Now , As the value of x increases , the difference d2 – d1 approaches zero .

    Therefore , When x\rightarrow \infty , the waves are in phase and interfere constructively

    Hence , They produce constructive interference

    Now , point P is moved along the x axis toward S1

    (a) The phase difference is zero at x\rightarrow \infty

    Therefore , the phase difference will increase as we move point P toward S1

    Hence , the phase difference between the waves will increase

    (b) Given , Phase difference \Phi = 0.50 \lambda = 0.50 * 2 = 1 m { wavelength , \lambda = 2.00 m }

    \Phi = d_{2}-d_{1}

    \Phi =\sqrt{(10)^{2}+(x)^{2}}-x ………………….. from equation (1)

    1 =\sqrt{(10)^{2}+(x)^{2}}-x

    1 + x =\sqrt{(10)^{2}+(x)^{2}}

    Squaring both sides , we get :

    (1 + x) ^{2}={(10)^{2}+(x)^{2}}

    1 + x ^{2}+2x=100+x^{2}

    1 +2x=100

    2x=100-1

    2x=99

    x=\frac{99}{2}=49.5 m

    Hence , the distance x = 49.5 m

    (c) Given , Phase difference \Phi = 1.00 \lambda = 1.00 * 2 = 2 m

    \Phi =\sqrt{(10)^{2}+(x)^{2}}-x

    2=\sqrt{(10)^{2}+(x)^{2}}-x

    2+x=\sqrt{(10)^{2}+(x)^{2}}

    Squaring both sides , we get :

    (2 + x) ^{2}={(10)^{2}+(x)^{2}}

    4 + x ^{2}+4x=100+x^{2}

    4 +4x=100

    4x=100-4

    4x=96

    x=\frac{96}{4}= 24 m

    Hence , the distance x = 24 m

    (d) Given , Phase difference \Phi = 1.50 \lambda = 1.50 * 2 = 3 m

    \Phi =\sqrt{(10)^{2}+(x)^{2}}-x

    3 =\sqrt{(10)^{2}+(x)^{2}}-x

    3 +x=\sqrt{(10)^{2}+(x)^{2}}

    Squaring both sides , we get :

    (3 + x) ^{2}={(10)^{2}+(x)^{2}}

    9 + x ^{2}+6x=100+x^{2}

    9+6x=100

    6x=100-9

    6x=91

    x=\frac{91}{6}= 15.2 m

    Hence , the distance x = 15.2 m

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