SATHEE: Wave Motion 5 Question 24

Wave Motion 5 Question 24

24. Two radio stations broadcast their programmes at the same amplitude $A$ and at slightly different frequencies $ω_{1}$ and $ω_{2}$ respectively, where $ω_{1} - ω_{2} = 10^{3} H z$ . A detector receives the signals from the two stations simultaneously. It can only detect signals of intensity $\geq 2 A^{2}$ .

(1993, 4M)

(a) Find the time interval between successive maxima of the intensity of the signal received by the detector. (b) Find the time for which the detector remains idle in each cycle of the intensity of the signal.

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Solution:

(a) If the detector is at $x = 0$ , the two radiowaves can be represented as

$y_{1} = A \sin ω_{1} t and y_{2} = A \sin ω_{2} t$

(Given, $A_{1} = A_{2} = A$ )

By the principle of superposition

$\begin{aligned} y = & y_{1} + y_{2} = A \sin ω_{1} t + A \sin ω_{2} t \\ y & = 2 A \cos (\frac{ω_{1} - ω_{2}}{2} t) \sin (\frac{ω_{1} + ω_{2}}{2} t) \\ = A_{0} \sin (\frac{ω_{1} + ω_{2}}{2} t) \end{aligned}$

Here, $A_{0} = 2 A \cos (\frac{ω_{1} - ω_{2}}{2} t)$

Since, $I \propto {(A_{0})}^{2} \propto 4 A^{2} \cos^{2} (\frac{ω_{1} - ω_{2}}{2} t)$

So, intensity will be maximum when

$\cos^{2} (\frac{ω_{1} - ω_{2}}{2} t) = maximum = 1$

$\begin{aligned} or \cos (\frac{ω_{1} - ω_{2}}{2} t) = \pm 1 \\ or \frac{ω_{1} - ω_{2}}{2} t = 0, π, 2 π \dots \\ i.e., t = 0, \frac{2 π}{ω_{1} - ω_{2}}, \frac{4 π}{ω_{1} - ω_{2}}, \dots \frac{2 n π}{ω_{1} - ω_{2}} n = 0, 1, 2 \dots \end{aligned}$

Therefore, time interval between any two successive maxima is $\frac{2 π}{ω_{1} - ω_{2}} = \frac{2 π}{10^{3}} s$ or $6.28 \times 10^{- 3} s$ .

(b) The detector can detect if resultant intensity $\geq 2 A^{2}$ , or the resultant amplitude $\geq \sqrt{2} A$ .

Hence, $2 A \cos (\frac{ω_{1} - ω_{2}}{2} t) \geq \sqrt{2} A$

$\cos (\frac{ω_{1} - ω_{2}}{2} t) \geq \frac{1}{\sqrt{2}}$

Therefore, the detector lies idle. When value of $\cos (\frac{ω_{1} - ω_{2}}{2} t)$ is between 0 and $1 / \sqrt{2}$ ]

or when $\frac{ω_{1} - ω_{2}}{2} t$ is between $\frac{π}{2}$ and $\frac{π}{4}$

or $t$ lies between

$\begin{aligned} \frac{π}{ω_{1} - ω_{2}} and \frac{π}{2 (ω_{1} - ω_{2})} \\ ∴ t & = \frac{π}{ω_{1} - ω_{2}} - \frac{π}{2 (ω_{1} - ω_{2})} \\ = \frac{π}{2 (ω_{1} - ω_{2})} = \frac{π}{2 \times 10^{3}} \\ t & = 1.57 \times 10^{- 3} s \end{aligned}$