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Sholpan [36]
3 years ago
12

Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus

a cycle. For example, if two waves have a phase difference of \frac{\pi}{4}, the interference effects would be the same as if the two waves had a phase difference of \frac{\pi}{4} + 2\pi. The complete criterion for constructive interference between two waves is therefore written as follows:
{\rm phase\ difference} = 0 + 2 \pi n \qquad {\rm for\ any\ integer}\ n
Write the full criterion for destructive interference between two waves.

Physics
1 answer:
poizon [28]3 years ago
3 0

Answer:

For destructive interference phase difference is

(2n+1)\pi where n∈ Whole numbers

Explanation:

For sinusoidal wave the interference affects the resultant intensity of the waves.

In the given example we have two waves interfering at a phase difference of \frac{\pi}{4} would lead to a constructive interference giving maximum amplitude at at the RMS value of the amplitude in resultant.

Also the effect is same as having a phase difference of  ( \frac{\pi}{4} + 2\pi) because after each 2π the waves repeat itself.

<em>In case of destructive interference the waves will be out of phase i.e. the amplitude vectors will be equally opposite in the direction at the same place on the same time as shown in figure.</em>

They have a phase difference of \pi or which is same as (2\pi+\pi)

Generalizing to:

a phase difference of (2n+1)\pi where n∈ {W}

{W}= set of whole numbers.

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Answer:

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Explanation:

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We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

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v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

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2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

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the electric fiel on the first one is:

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then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

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