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Question

Two superimposing waves are represented by the equations x1=5sin2π(20t0.1x) and x2=10sin2π(20t0.2x) where x and amplitude of each wave are given in meters and t is in seconds. Find the ratio of maximum intensity to minimum intensity.

A
2:1
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B
1:2
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C
9:1
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D
1:9
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Solution

The correct option is C 9:1
Given that,
Amplitude of wave 1, (A1)=5 m
Amplitude of wave 2, (A2)=10 m

and we know that,

Intensity (I)(Amplitude(A))2

Let I1 and I2 be the intensities of wave1 and wave 2
Then,
Maximum intensity is given by
Imax(A1+A2)2

Minimum intensity is given by
Imin(A1A2)2.

From this, ImaxImin=(A1+A2)(A1A2)2

From the given data,

ImaxImin=(5+10)(510)2
ImaxImin=(155)2=91
Imax:Imin=9:1

Hence, option (c) is the correct answer.

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