Waves Practice Questions

$f^{'}=f_0 [\;\frac{v-v_0}{v-v_s}]$ $\Rightarrow f^{'}=f_0 [\;\frac{v+\frac{v}{5}}{v}]$ $\Rightarrow f^{'}=\;\frac{6 f_0}{5}$ $\Rightarrow \% \;\text{ change }\;=20$

$A_{n e t}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \varphi }$ $\sqrt{3} A=\sqrt{A^{2}+A^{2}+2 A^{2} \cos \varphi }$ $3 A^{2}=2 A^{2}+2 A^{2} \cos \varphi$ $\cos \varphi =\;\frac{1}{2}$ $\varphi =60^{\circ}$

First overtone of open pipe $=\;\frac{V_{2}}{L_{2}}$ First overtone of closed pipe at one end $=\;\frac{3 {v}}{4 L}$ As per question, $\;\frac{3 V}{4 L}=\;\frac{V_{2}}{L}$ $\Rightarrow \;\; \sqrt{\;\frac{B}{\rho _{1}}} \cdot \;\frac{3}{4 L}=\sqrt{\;\frac{B}{\rho _{2}}} \cdot \;\frac{1}{L_{2}}$ $( \because V=\sqrt{\;\frac{B}{\rho }})$ $\Rightarrow \;\; L_{2}=\;\frac{4 L}{3} \sqrt{\;\frac{\rho _{1}}{\rho _{2}}}$....(i) According to question, the length of the open pipe is $\;\frac{x}{3} L \sqrt{\;\frac{\rho _{1}}{\rho _{2}}}$....(ii) Comparing Eqs. (i) and (ii), we get $x=4$

Given, mass per unit length, $\mu =0.135 {g} / {cm}$ Transverse wave equation, $y=-0.21 \sin (x+30 t)$ From given equation, $\omega =30 {rad} / {s}, k=1$ Speed of wave, $v=\;\frac{\omega }{k}=\;\frac{30}{1}=30 {ms}^{-1}$ Also, $\;\; v=\sqrt{\;\frac{T}{\mu }}$ $\Rightarrow \;\; T=v^{2} \mu$ $T=(30)^{2} \times \;\frac{0.135 \times 10^{-3}}{10^{-2}}$ $\;\; =900 \times 0.0135$ $\;\; =12.15 {N}$ $\;\; =1215 \times 10^{-2} {N}$ Hence, $x=1215$

Given, velocity of car X, $\text{v}_{\text{x}}=36 {\text{kmh}}^{-1}=36 \times \frac{5}{18} {\text{ms}}^{-1}=10 {\text{ms}}^{-1}$ Velocity of car $\text{Y}, \text{v}_{\text{y}}=72 {\text{kmh}}^{-1}$ $=72 \times \frac{5}{18} {\text{ms}}^{-1}=20 {\text{ms}}^{-1}$ Frequency of sound heard by passenger in car Y emitted by passenger in car X, f′ =1320 Hz Velocity of sound, $v_s$ = 340 m/s. By Doppler's effect, $\text{f}'={\text{f}(\text{v}_{\text{s}} ±v_{\text{O}})}/{(v_{s} \∓v_{s})}$ Velocity of observer is equal to velocity of car, $\text{v}_{\text{o}}=20 {\text{ms}}^{-1}$ Velocity of source is equal to velocity of car X, $\text{v}_{\text{s}}=10 {\text{ms}}^{-1}$ Substituting the values in above expression, we get $1320=\frac{\text{f}(340+20)}{(340-10)}$ $\Rightarrow 1320=\frac{360 \text{f}}{330}$ $\Rightarrow \text{f}=1320 \times \frac{11}{12}=1210 {\text{Hz}}$ Thus, the actual frequency is 1210 Hz.

Given, linear mass density, $\lambda=9 \times 10^{-4} {\text{kg}} / {\text{m}}$ Tension, $\text{T}'=900 {\text{N}}$ Resonance frequency, $\text{f}_{1}=500 {\text{Hz}}$ and $\text{f}_{2}=550 {\text{Hz}}$ Let, length of wire be l. As we know that,$\text{f}_{\text{n}}={\text{n}}{2 \text{I}} \\sqrt{\frac{\text{T}}{\lambda}}$ $\Rightarrow 500=\frac{n}{2 l} \\sqrt{\frac{900}{9 \times 10^{-4}}}=\frac{n}{2 \text{l}} \\sqrt{10^{6}}$ $\Rightarrow 500=\frac{n}{2 l} \times 10^{3}$ $\Rightarrow \frac{n}{2 \text{l}}=\frac{500}{1000}=\frac{1}{2}$ $\Rightarrow \frac{n}{\text{l}}=1$ ⇒ n = l ...(i) and$550=\frac{n+1}{2 I} \\sqrt{\frac{9 \times 10^{6}}{9}}$ $\Rightarrow 550=\frac{n+1}{2 l} \times 10^{3}$ $\Rightarrow 550={\text{I}+1}{2 \text{I}} \times 10^{3}$ [From Eq, (i)] $\Rightarrow \frac{55}{100}=\frac{\text{I}+1}{2 \text{I}}$ $\Rightarrow \frac{11}{20}=\frac{\text{I}+1}{2 \text{I}}$ $\Rightarrow 22 l=20 l+20$ $\Rightarrow 2 \text{l}=20$ ⇒ l = 10 m

Wall as an observer Frequency received by wall ${f}_{1}=f_{0}(\frac{{C}}{{C}-{V}})$ Again wall as a source Frequency received by observer on car ${f}_{2}={f}_{1}(\frac{{C}+{V}}{{C}})$ ${f}_{2}={f}_{0}(\frac{{C}+{V}}{{C}-{V}})$ $500=400(\frac{{C}+{V}}{{C}-{V}})$ $\frac{5}{4}=\frac{{C}+{V}}{{C}-{V}}$ ${C}=9 {V}$ ${V}=\frac{{C}}{9}=\frac{330}{9} {m} / {s}$ ${V}=\frac{330}{9} \\times \frac{18}{5}=132$ km / hr

The equation of a travelling wave in standard form, $y=A \sin (\omega t ± k x)$ Only option (a), i.e. $y=A \sin (15 x-2 t)$ satisfies this equation.

At ${t}=0, {y}=\frac{1}{1+{x}^{2}}$ At time ${t}={t}, {y}=\frac{1}{1+({x}-{vt})^{2}}$ At $t=1, y=\frac{1}{1+( x-v)^{2}}$ .......(i) At ${t}=1, {y}=\frac{1}{1+({x}-2)^{2}}$ ........(ii) Comparing (i) & (ii) ${v}=2 {m} / {s}$

${x}={A} \sin \omega {t}+{B} \cos \omega {t}$ ${v}=\frac{{dx}}{{dt}}={A} \omega \cos \omega {t}-{B} \omega \sin \omega {t}$ At ${t}=0, {x}(0)={B}$ ${v}(0)={A} \omega$ ${x}={A} \sin \omega {t}+{B} \sin (\omega {t}+90°)$ $A_{\text{net }}=\\sqrt{A^{2}+B^{2}}$ $\tan \alpha =\frac{B}{A} \Rightarrow\cot \alpha =\frac{A}{B}$ $\Rightarrowx=\\sqrt{A^{2}+B^{2}} \sin (\omega t+\alpha )$ $\Rightarrowx=\\sqrt{A^{2}+B^{2}} \cos (\omega t-(90-\alpha ))$ $x=C \cos (\omega t-\phi )$ $\RightarrowC=\\sqrt{A^{2}+B^{2}}$ $C=\\sqrt{\frac{[v(0)]^{2}}{\omega ^{2}}+[x(0)]^{2}}$ $\phi =90-\alpha$ $\tan \alpha =\cos \alpha =\frac{A}{B}$ $\Rightarrow \tan \phi =\frac{v(0)}{x(0) \.\omega }$ $\phi =\tan ^{-1}(\frac{{v}(0)}{{x}(0) \omega })$

Error in tension is $4 \%$. Speed of transverse wave, $v=\sqrt{\;\frac{\;\text{ Tension }\;(T)}{\;\text{ Mass per unit length }\;(\mu )}}$ On squaring both side $\Rightarrow \;\; v^{2}=\;\frac{T}{\mu }$ By using the concept of % error calculation $\Rightarrow \;\; \;\frac{2 \Delta v}{v}=\;\frac{\Delta T}{T} \Rightarrow \;\frac{\Delta v}{v}=\;\frac{1}{2} \;\frac{\Delta T}{T}$ $\therefore \;\; \;\frac{\Delta v}{v} \times 100=\;\frac{1}{2} \;\frac{\Delta T}{T} \times 100=\;\frac{4}{2}=2 \%$

Given, diameter of the column tube, $d=6 {cm}=6 \times 10^{-2} {m}$ Frequency of tuning fork, $f=504 {Hz}$ Speed of sound at given temperature, $v=336 {ms}^{-1}$ As, this is a closed organ pipe. Let $L$ be the length of tube and $\lambda$ be the wavelength, thenand$L+ {e} \;\; =\lambda / 4=v / 4 f \;\; ( \because \lambda =\;\frac{v}{f})$ ${e} \;\; =0.6 \times d / 2$ $\;\; =\;\frac{6}{10} \times \;\frac{6 \times 10^{-2}}{2}=0.018$ $\Rightarrow \;\; L+ {e} \;\; =\;\frac{v}{4 f}=\;\frac{336}{4 \times 504}$ $\Rightarrow \;\; L \;\; =\;\frac{336}{2016}-0.018=0.1667-0.018$ $\;\; =0.1487 {m}=14.87 {cm}$ $\;∼eq 14.8 {cm}$

Given, equation of wave of first string, $y_{1}=\text{A}_{1} \sin k(x-v t)$ Equation of wave of second string, $y_{2}=\text{A}_{2} \sin k(x-v t+x_{0})$ $\text{A}_{1}=12 {\text{mm}}, \text{A}_{2}=5 {\text{mm}}, x_{0}=3.5 {\text{cm}}$ and $k=6.28 {\text{cm}}^{-1}$ The equation of waves after substitution of values can beexpressed as $y_{1}=12 \sin 6.28(x-v t)$ and $y_{2}=5 \sin 6.28(x-v t+3.5)$ The phase difference between two waves can be calculated as $\Delta\phi =\frac{2 \pi }{\lambda} \Deltax$ From equation of waves, the value of $\Deltax=(x-v t+3.5)-(x-v t)=3.5 {\text{cm}}$ We know that, wave number is k = 2 π / λ. Now, the phase difference is ∆ϕ = k∆x= 6.28 × 3.5 = 2 π × 3.5 = 7 π The amplitude of resulting wave can be calculated by using thefollowing relation, $\text{A}_{{\text{net} }}=\\sqrt{\text{A}_{1}^{2}+\text{A}_{2}^{2}+2 \text{A}_{1} \text{A}_{2} \cos \Delta\phi }$ Substituting the values in above expression, $\text{A}_{{\text{net} }}=\\sqrt{12^{2}+5^{2}+2 \times 12 \times 5 \times \cos 7 \pi }$ $=\\sqrt{144+25+120 \times (-1)}$ $\text{A}_{{\text{net} }}=\\sqrt{49}=7 {\text{mm}}$ Thus, the amplitude of resultant wave is 7 mm.

Given, $v_{A}=v_{B}=7.2 {kmh}^{-1}$ $=\;\frac{72}{10} \times \;\frac{5}{18}=2 {ms}^{-1}$ Frequency of source, $f_{ {s}}=676 {Hz}$ Speed of sound in air, $v=340 {ms}^{-1}$ Let $f_{0}$ be the frequency heard by each driver. By using Doppler effect for $A$, $(v-v_{A}) f_{s} \;\; =(v+v_{B}) f_{0}$ $\Rightarrow f_{0}=(\;\frac{v+v_{A}}{v-v_{B}}) f_{s} \;\; =(\;\frac{340+2}{340-2}) 676=\;\frac{342}{338} \times 676=684 {Hz}$ Now, beat frequency $=f_{0}-f_{s}=684-676=8 {Hz}$

As source and detector move away from each other as shownbelow $\text{table} {←}_{\text{Source}},, {\to } _{\text{Detector}}$ $v_{s}=$ velocity of source with respect to ground = 20 m/s $v_{d}=$ velocity of detector with respect to ground = 20 m/s According to Doppler's effect, $\text{f}_{{\text{apparent} }}={\text{f}(v_{{\text{sound} }}-v_{d})}/{(v_{{\text{sound} }}+v_{s})}$ $\Rightarrow 1800=\frac{\text{f}(340-20)}{(340+20)}$ $1800=\text{f}(\frac{320}{360})$ $\text{f}=\frac{1800 \times 36}{32}$ $=\frac{1800 \times 9}{8}$ $=\frac{16200}{8}$ $\Rightarrow \text{f}=2025 {\text{Hz}}$

Given, $f=340 {Hz}$ Let beat frequency of tuning fork $A$ is $f_{A}$ and $f_{A}^{'}$ be the frequency of tuning fork $A$ after filling. In first case, beat frequency $=5$ $\Rightarrow \;f_{A}=340 ± 5$ $\Rightarrow \;f_{A}=345 {Hz} \;\text{ or }\; 335 {Hz}$ After filling tuning fork $A, f_{A}^{'}=340 ± 2$ $\Rightarrow \;\; f_{A}^{'}=342 \;\text{ or }\; 338 {Hz}$ As on filling, frequency of $A$ increases. Hence, original frequency of tuning fork $A$ is $335 {Hz}$.

For a closed organ pipe, the shortest length ($L$) corresponds to the fundamental frequency, where the length is one-quarter of the wavelength ($\lambda$).
Given frequency $f = 250 {\text{ Hz}}$ and speed of sound $v = 340 {\text{ m/s}}$.
The relationship between speed, frequency, and wavelength is $v = f\lambda$.
For the fundamental mode in a closed organ pipe:
$L = \frac{\lambda}{4}$
Substitute $\lambda = \frac{v}{f}$ into the equation for $L$:
$L = \frac{v}{4f}$
Plugging in the values:
$L = \frac{340 {\text{ m/s}}}{4 \times 250 {\text{ Hz}}} = \frac{340}{1000} {\text{ m}} = 0.34 {\text{ m}}$
To convert meters to centimeters:
$L = 0.34 \times 100 {\text{ cm}} = 34 {\text{ cm}}$
[CORRECT_OPTION: N/A]

Given, The frequency of the sound wave, $f=245 {Hz}$ The speed of the travelling wave, $v=300 {m} / {s}$ As, total distance of to and fro motion is $6 {cm}$. Hence, the amplitude of the wave, $A=6 / 2=3 {cm}=0.03 {m}$ As we know, Angular frequency is given as $\omega =2 \pi f$ Substituting the value of $f$, we get $\omega =2 \pi (245)$ $\omega =1.54 \times 10^{3} {rad} / {s}$ Propagation constant is given as $k=\;\frac{\omega }{v}$ Substituting the value of $v$ and $\omega$, we get $k=\;\frac{1.54 \times 10^{3}}{300} \Rightarrow k=5.1 {m}^{-1}$ General mathematical expression for a travelling wave is given as $y=A \sin (k x-\omega t)$ Substituting the values in the above equation, we get $y=0.03 \sin (5.1 x-1.5 \times 10^{3} t)$