Waves Practice Questions

$v_s=\sqrt{{\gamma P}/ \rho }$ $v_\text{gas}/v_\text{air}=\sqrt{\rho _\text{air}/\rho _\text{gas}}$ $\Rightarrow v_\text{gas}/300=1/\sqrt2$ $\therefore v_\text{gas}=150\sqrt2$ Now n_2-n_1=v_\text\frac{gas}{2l} $=\frac{150\sqrt2}{2(1)}=75\sqrt2$ $\Deltan=106.06\text{Hz}$

Let amplitude of each wave is A.Resultant wave equation$=A\sin\omega t+A\sin(\omega t-\pi /4)$ $+A\sin(\omega t+\pi /4)$ $=(\sqrt2+1)A\sin\omega t$Resultant wave amplitude $=(\sqrt2+1)A$as $I \propto A^2$so, $I/I_0=(\sqrt2+1)^2$ $I=5.8I_0$

$\Delta{p}=\ {BkS}_{0}$ $=\ \rho {v}^{2}\times \frac{\omega }{ {v}} \times {S}_{0}$ $\Rightarrow {S}_{0}= \frac{\Delta{p}}{\ \rho {v} \omega }$ $\approx \frac{10}{1 \\times 300 \\times 1000} \ {m}$ $= \frac{1}{30} \ {mm} \approx \frac{3}{100} \ {mm}$

Given $\ {T}$ to $\ {C}\ 1.5 \ {m}$ $\ {C}$ to $\ {C}\ 5 \ {m}$ $\ {T}$ to $\ {C}= (2 {n}_{1}+1 ) \frac{\lambda}{2}$ $\ {C}$ to $\ {C}=\ {n}_{2} \lambda$ \ \frac{1.5}{5}=\frac{(2 n_{1}+1 )}{2 n_{2}} \ $\Rightarrow 3 n_{2}=10 n_{1}+5$ $\ {n}_{1}=1, \ {n}_{2}=5 \to \lambda=1$ $\ {n}_{1}=4, \ {n}_{2}=15 \to \lambda=1 / 3$ $\ {n}_{1}=7, \ {n}_{2}=25 \to \lambda=1 / 5$

$v_1=(c/{c-v})v_0$ $v_2=(c/{c+v})v_0$ beat frequency $= v_1-v_2$ $=cv_0(\frac{c+v-c+v}{c^{2-}v^2}) =\frac{2cv_0^2v}{c^{2-}v^2}$ $\approx {2cv_0v}/c^2={2v_0v}/c=2$ $\Rightarrow {2\times 1400\times v}/350=2$ $\Rightarrow v=1/4 m/s$

$f_{1}=(\frac{330}{330-v_{B}} ) 420$ $f_{2}=(\frac{330+v_{0}}{330})$ $(\frac{330}{330-v_{B}} ) 420$ $490=(\frac{330+v_{B}}{330-v_{B}} ) 420$ $\frac{7}{6}= \frac{330+v_{B}}{330-v_{B}}$ $v_{B}= \frac{330}{13} \ {m} / \ {s}$ $= \frac{330}{13} \times \frac{18}{5} \approx 91 \ {km} / {hr}$

${f}_{1}=$ frequency heard by wall $= {f}_{ {s}}=({{v}_{ {s}}}/{ {v}_{ {s}}- {v}_{ {c}}} )$ ${f}_{2}=$ frequency heard by driver after reflection from wall ${f}_{2}=({{v}_{ {s}}+ {v}_{ {c}}}/{ {v}_{ {s}}} ) {f}_{1}$ $=({{v}_{ {s}}+ {v}_{ {c}}}/{ {v}_{ {s}}- {v}_{ {c}}} ) {f}_{0}$ $\frac{{f}_{2}}{ {f}_{0}}={{v}_{ {s}}- {v}_{ {c}}}/{ {v}_{ {s}}+ {v}_{ {c}}}$ $\frac{48}{44}= {{v}_{ {s}}- {v}_{ {c}}}/{ {v}_{ {s}}+ {v}_{ {c}}}$ $12({v}_{ {s}}+ {v}_{{c}} )=11({v}_{ {s}}- {v}_{ {c}} )$ $23 {v}_{ {c}}= {v}_{ {s}}$ ${v}_{ {c}}={{v}_{ {s}}}/{23}= \frac{345}{23}=15 { m} / { s}$ $= \frac{15 \\times 18}{5}=54{km} / {hr}$

${\text{f}}_{{\text{observed} }} \Rightarrow ({{v}_{ {\text{sound} }}}/{{v}_{{\text{sound} }}-{v} \cos \theta }) {\text{f}}_{0}$ initially $\theta$ will be less $\Rightarrow \cos \theta$ more $\therefore {\text{f}}_{{\text{observed} }}$ more, then it will decrease.

Velocity of transverse wave V $\propto \text{sqrtT}$ $V\to V/2\Rightarrow T\to T'=T/4$ $T'={2.06\times 10^4}/4=5.15\times 10^3N$

$\Rightarrow \lambda=2 (l_{2}-l_{1}\ ) \Rightarrow 2 \\times (24.5-17)$ $\Rightarrow2 \\times 7.5=15 \ {cm}$ $\& \ {v}=\ {f} \lambda\Rightarrow 330=\lambda\times 15 \\times 10^{-2}$ \lambda= \frac{330}{15} \\times 100 \ $\Rightarrow \frac{1100 \\times 100}{5}$ $\Rightarrow2200 \ {Hz}$

$\frac{nv}{2l}=420$ $\frac{(n+1)v}{2l}=490$ $v/{2l}=70$ $l=v/140=1/140\sqrt{540/{6\times 10^{-3}}}=2.142$

${V} \propto \lambda \;\;\;{T}_{2}=8 {g}$ ${T}_{1}=2 {g}$ $\frac{V_{1}}{V_{2}}= \frac{\lambda_{1}}{\lambda_{2}}$ $\lambda_{2}=\frac{{V}_{2}}{ {V}_{1}} \lambda_{1}$ $=\sqrt{\frac{{T}_{2}}{{T}_{1}}} \times \lambda_{ {I}}$ $=\sqrt{\frac{8 {g}}{2 {g}}} \lambda_{1}$ $=2 \\times 6=12 \ {cm}$

${f}= \frac{1}{2 \text{l}} \sqrt{\frac{{T}}{\mu}}$ For identical string $l$ and $\mu$ will be same $\ {f} \propto \sqrt{{T}}$ $\frac{450}{300}=\sqrt{{{T}_{ {x}}}/{ {T}_{ {y}}}}$ ${{T}_{ {x}}}/{ {T}_{ {y}}}= \frac{9}{4}=2.25$

$480=500\ ( {300+ {\text{V}}_{ {o}}}/{300} )$ $\frac{24}{25}=1+ \frac{\text{V}_{0}}{300}$ $\ {\text{V}_{0}} /{300}=\ \frac{1}{25}$ $\ |V_{0} | =12$ $530=500\ (\ \frac{300+\ {V}_{0}}{300} )$ $\ \frac{53}{50}=1+\ {\ {\text{V}}_{ {\text{o}}}}/{300}$ $\ {\text{V}_{\text{o}}}/{300}= \frac{-3}{50}$ $\ |V_{0} |=18$Hence Velocities -12,18

$f^{\1}= (\ \frac{V-u}{V}\ ) f,f^{x}=\ (\ \frac{V+u}{V}\ ) f$ $f^{x}-f^{ {1}} =\ (\ \frac{V+u-V+u}{V}\ ) f= \frac{2 u f}{V}$ $\Rightarrow 10=\ \frac{2 u \times 660}{330}$ $u=\ \frac{10}{4}=2.5$ m / s

Velocity of wave on string V = $\sqrt{T/\mu }$ = $\sqrt{8/5\times 1000}$ = 40 m/s Now, wavelength of wave λ = $v/n$ = $\frac{40}{100}$ m Separation b/w successive nodes , $\lambda /2$ = $\frac{20}{100}$ m = 20 cm

$\beta =10 \log (\ \frac{I}{I_{0}}\ ) 120$ $=10 \log \ (\ \frac{I}{I_{0}}\ ) 10^{12}= \frac{I}{I_{0}}$ $I=1=\ \frac{\rho }{4 \pi d^{2}} d^{2}$ $=\ \frac{2}{4 \pi } d=\ \frac{1}{\\sqrt{2 \pi }}$ $=0.3989 \;m d \approx 40 {cm}$

Given string is in $3^{ { rd }}$ harmonic mode$\Rightarrown=\ \frac{3 v}{2 l}$ $240=\ \frac{3(v)}{2(2)}$v = 320 m / sFundamental frequency $\Rightarrow n=\ \frac{v}{2 l} \Rightarrow\frac{320}{2(2)}=80 H z$