# Two radio stations broadcast their program’s at the same amplitude A and at slightly frequencies ω1 and ω2 respectively, where ω1 – ω2 = 103 Hz A detector receives the signals from the two stations simultaneously. It can only detect signals of intensity ≥ 2A2   (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.

10 years ago
Hello Student,
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 – ω2 or, 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 – ω2 and 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