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??

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

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over

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