Waves

$7.2*\frac{5}{18}=2m/s$ [V_A = V_B = 2m/s]

$f{\text{'}}_B=f_A\left(\frac{v+v_B}{v-v_A}\right)$

$\downarrow$     

App freq heard by car B

$f{\text{'}}_B=676\left[\frac{340+2}{340-2}\right]=676*\frac{342}{338}=684Hz$  

Similarly, $f{\text{'}}_A=f_B\left(\frac{v+v_B}{v-v_B}\right)=684Hz$

$\therefore$ Beat frequency heard by both = 684 – 676 = 8Hz

Travelling wave is represented by $f\left(t\pm \frac{x}{v}\right)$ $\frac{}{}$

Thus, y = A sin (15x – 2t) represents a travelling wave

For first resonance,

$\mathcal{l}+0.3d=\frac{v}{4f}$

$\Rightarrow \mathcal{l}+0.3*6=\frac{336*100}{4*504}\Rightarrow \mathcal{l}=14.8cm$

Initially wall will act as observer, so with the help of Doppler's effect, we can write

$f\text{'}=\frac{c}{c-v_0}{f}_0$  

Now wall will act as source of sound of frequency f', so With the help of Doppler's effect, we can write

$f\text{'}\text{'}=\frac{c+v_0}{c}{f}^{\text{'}}=\frac{c+v_0}{c-v_0}{f}_0$              

$\Rightarrow 500=\left(\frac{c+v_0}{c-v_0}\right)400\Rightarrow 5c-5v_0=4c+4v_0\Rightarrow v_0=\frac{c}{9}=\frac{330}{9}m/s=\frac{330}{9}*\frac{18}{5}km/h=132km/h$

$f^1=f_0*\left[\frac{340+20}{340-10}\right]=1320$

=> f₀ = 1210 Hz

Phase difference between the two waves,

$?=k{x}_0=6.28*3.5=7\pi rad$              

Amplitude of resulting wave,

$A=\sqrt[]{A_1^2+A_2^2+2A_1{A}_{2}cos?}$              

$=\sqrt[]{12^2+5^2+2*12*5*cos\left(7\pi \right)}$              

$kx=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$\Rightarrow 1.57x=\frac{\pi }{2}=\frac{3.14}{2}=1.57$

x = 1 cm

$f_1=f_o\left(\frac{v}{v-v_s}\right)\mathrm{}{f}_1-f_2=2$

$f_2=f_o\left(\frac{v}{v+v_s}\right)\mathrm{}{f}_{0}v\left[\frac{v+v_s-\left(v-v_s\right)}{v^2-v_s^2}\right]=2;\mathrm{}\frac{2f_{0}v{v}_s}{v^2-v_s^2}=2$

$?{\mathrm{v}}_{\mathrm{s}}?\mathrm{v}\mathrm{}\therefore {\mathrm{v}}_{\mathrm{s}}=\frac{1}{4}\mathrm{m}/\mathrm{s}$

$y=0.5sin\frac{2\pi }{\lambda }\left(400t-x\right)m$

$y=y_{0}sin\left(\underset{\omega }{\underset{?}{\frac{2\pi }{\lambda }*400}}\underset{k}{\underset{?}{t-\frac{2\pi }{\lambda }}}x\right)m$

Wave speed, $V=\frac{\omega }{k}=400m/s$

$\frac{nv}{0.6}=400\&\frac{\left(n+1\right)v}{0.6}=450$

$\Rightarrow \left[\frac{0.6*450}{v}+1\right]$ $\frac{v}{0.6}=450$

v = 30

$\Rightarrow \sqrt[]{\frac{T}{\mu }}=30$

^{$\Rightarrow \frac{2700}{\mu }=900$}

^{$\mu =3$}