Two radio stations broadcast their program’s at the same amplitude A and at slightly frequencies ω 1 and ω 2 respectively, where ω 1 – ω 2 = 10 3 Hz A detector receives the signals from the two stations simultaneously. It can only detect signals of intensity ≥ 2A 2 (i) Find the time interval between successive maxima of the intensity of the signal received by the detector. (ii) Find the time for which the detector remains idle in each cycle of the intensity of the signal.

Question

Aditi Chauhan · Answer

Hello Student,

Please find the answer to your question

Let the two radio waves be represented by the equation

y₁ = A sin 2πv₁ t

y₂ = A sin 2πv₂ t

The equation of resultant wave according to superposition principle

y = y₁ + y₂ = a sin 2πv₁ t + A sin 2πv₂ t

= A [sin 2 πv₁ t + sin 2 πv₂ t]

= A x 2 sin (2πv₁ + 2 πv₂) t/2 cos (2 πv₁ + 2 πv₂ )t/2

= 2A sin π(v₁ + v₂) t cos π(v₁ – v₂) t

Where the amplitude A’ = 2A cos π (v₁ – v₂) t

Now, intensity ∝ (Amplitude)²

⇒ I∝ A’²

⇒ I∝ 4A² cos² π (v₁ + v₂) t

The intensity will be maximum when

Cos² π (v₁ – v₂) t = 1

Or, cos π (v₁ + v₂) t = 1

Or, π v₁ – v₂ t nπ

⇒ (ω₁ – ω₂)/2 t = nπ or, t = 2nπ/ ω₁ – ω₂

∴ Time interval between two maxima

Or, 2nπ/ ω₁ – ω₂- 2(n – 1)π/ ω₁ – ω₂or, 2π ω₁ – ω₂= 2π/10³ sec

Time interval between two successive maximas is

2π x 10^-3 sec

(ii) For the detector to sense the radio waves, the resultant intensity ≥ 2 A²

∴ Resultant amplitude ≥ √2 A

Or, 2 A cos π (v₁ – v₂) t ≥ √2A

Or, cos π (v₁ – v₂) t ≥ 1/√2 or, cos [(ω₁ – ω₂) t/2] ≥ 1/√2

The detector lies idle when the values of cos[(ω₁ – ω₂) t/2] is between 0 and 1/√2

∴ (ω₁ – ω₂) t/2 is between π/2 and π/4

∴ t₁ = π/ ω₁ – ω₂and t₂ = π/2 (ω₁ – ω₂)

∴ The time gap = t₁ – t₂

= π/ ω₁ – ω₂ – π/2 (ω₁ – ω₂) = π/2 (ω₁ – ω₂)

= π/2 x 10^-3 sec.

Thanks
Aditi Chauhan
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