$\Rightarrow 10[\sin A+\sin B]$
where
$B=2 \pi(100 t-0.02 x)$ and
$A=2 \pi(100 t+0.02 x)$
Thus,
$\Rightarrow 10\left[2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}\right]$
$\Rightarrow 20 \sin (2 \pi 100 t) \sin (2 \pi 0.02 x)$
Comparing the above equation with standard standing wave equation, we get amplitude $=20$ and wave vector
$k=\frac{2 \pi}{\lambda}=2 \pi \times 0.02 \Rightarrow \lambda=50$
Therefore, the loop length $=\frac{\lambda}{2}=25$
Assume that the sound of the whistle is composed of components varying in frequency from $f_1=800 \mathrm{~Hz}$ to $f_2=1120 \mathrm{~Hz}$, as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus $320 \mathrm{~Hz}$. The speed of sound in still air is $340 \mathrm{~m} / \mathrm{s}$.
$1.$ The speed of sound of the whistle is
$(A)$ $340 \mathrm{~m} / \mathrm{s}$ for passengers in $A$ and $310 \mathrm{~m} / \mathrm{s}$ for passengers in $B$
$(B)$ $360 \mathrm{~m} / \mathrm{s}$ for passengers in $A$ and $310 \mathrm{~m} / \mathrm{s}$ for passengers in $B$
$(C)$ $310 \mathrm{~m} / \mathrm{s}$ for passengers in $A$ and $360 \mathrm{~m} / \mathrm{s}$ for passengers in $B$
$(D)$ $340 \mathrm{~m} / \mathrm{s}$ for passengers in both the trains
$2.$ The distribution of the sound intensity of the whistle as observed by the passengers in train $\mathrm{A}$ is best represented by
$Image$
$3.$ The spread of frequency as observed by the passengers in train $B$ is
$(A)$ $310 \mathrm{~Hz}$ $(B)$ $330 \mathrm{~Hz}$ $(C)$ $350 \mathrm{~Hz}$ $(D)$ $290 \mathrm{~Hz}$
Give the answer question $1,2$ and $3.$
${z_1} = a\cos (kx - \omega \,t)$.....$(A)$
${z_2} = a\cos (kx + \omega \,t)$.....$(B)$
${z_3} = a\cos (ky - \omega \,t)$..... $(C)$