Wave Motion 5 Question 24
24. Two radio stations broadcast their programmes at the same amplitude $A$ and at slightly different frequencies $\omega _1$ and $\omega _2$ respectively, where $\omega _1-\omega _2=10^{3} Hz$. 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.
Show Answer
Answer:
Correct Answer: 24. (a) $6.28 \times 10^{-3} \mathrm{~s}$ (b) $1.57 \times 10^{-3} \mathrm{~s}$
Solution:
- (a) If the detector is at $x=0$, the two radiowaves can be represented as
$$ y _1=A \sin \omega _1 t \text { and } y _2=A \sin \omega _2 t $$
(Given, $A _1=A _2=A$ )
By the principle of superposition
$$ \begin{aligned} y= & y _1+y _2=A \sin \omega _1 t+A \sin \omega _2 t \\ y & =2 A \cos \left(\frac{\omega _1-\omega _2}{2} t\right) \sin \left(\frac{\omega _1+\omega _2}{2} t\right) \\ & =A _0 \sin \left(\frac{\omega _1+\omega _2}{2} t\right) \end{aligned} $$
Here, $A _0=2 A \cos \left(\frac{\omega _1-\omega _2}{2} t\right)$
Since, $I \propto\left(A _0\right)^{2} \propto 4 A^{2} \cos ^{2}\left(\frac{\omega _1-\omega _2}{2} t\right)$
So, intensity will be maximum when
$$ \cos ^{2}\left(\frac{\omega _1-\omega _2}{2} t\right)=\text { maximum }=1 $$
$$ \begin{aligned} & \text { or } \quad \cos \left(\frac{\omega _1-\omega _2}{2} t\right)= \pm 1 \\ & \text { or } \quad \frac{\omega _1-\omega _2}{2} t=0, \pi, 2 \pi \ldots \\ & \text { i.e., } t=0, \frac{2 \pi}{\omega _1-\omega _2}, \frac{4 \pi}{\omega _1-\omega _2}, \ldots \frac{2 n \pi}{\omega _1-\omega _2} \quad n=0,1,2 \ldots \end{aligned} $$
Therefore, time interval between any two successive maxima is $\frac{2 \pi}{\omega _1-\omega _2}=\frac{2 \pi}{10^{3}} s$ 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, $2 A \cos \left(\frac{\omega _1-\omega _2}{2} t\right) \geq \sqrt{2} A$
$$ \cos \left(\frac{\omega _1-\omega _2}{2} t\right) \geq \frac{1}{\sqrt{2}} $$
Therefore, the detector lies idle. When value of $\cos \left(\frac{\omega _1-\omega _2}{2} t\right)$ is between 0 and $1 / \sqrt{2}$ ]
or when $\frac{\omega _1-\omega _2}{2} t$ is between $\frac{\pi}{2}$ and $\frac{\pi}{4}$
or $t$ lies between
$$ \begin{aligned} & \frac{\pi}{\omega _1-\omega _2} \text { and } \frac{\pi}{2\left(\omega _1-\omega _2\right)} \\ \therefore \quad t & =\frac{\pi}{\omega _1-\omega _2}-\frac{\pi}{2\left(\omega _1-\omega _2\right)} \\ & =\frac{\pi}{2\left(\omega _1-\omega _2\right)}=\frac{\pi}{2 \times 10^{3}} \\ t & =1.57 \times 10^{-3} s \end{aligned} $$
Hence, the detector lies idle for a time of $1.57 \times 10^{-3} s$ in each cycle.
