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JEE Main · Physics

Waves Practice Questions

Generate JEE Main level questions on Waves. Focus on Doppler effect, Beats, and Standing waves.

152 questions · 20 PYQs · 0 AI practice · JEE Main 2027

Q121JEE Main 2017MCQ

In an experiment to determine the period of a simple pendulum of length 1 m, it isattached to different spherical bobs of radii $r_1$ and $r_2$ . The two spherical bobs haveuniform mass distribution. If the relative difference in the periods, is found to be $5\times 10^{−4}s$ , the difference in radii, $|r_1−r_2|$ is best given by :[Main 9 April 2017]

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📖 Explanation

The period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L+r}{g}}$ , where $L=1 \text{ m}$ is the length of the string and $r$ is the radius of the bob.
Differentiating with respect to $r$ gives the relation for small changes: $\frac{dT}{T} = \frac{1}{2} \frac{dr}{L+r}$ .
Given the relative difference in periods $\frac{\Delta T}{T} = 5 \times 10^{-4}$ and approximating $L+r \approx L = 1 \text{ m}$ :
$5 \times 10^{-4} = \frac{1}{2} \frac{|r_1 - r_2|}{1 \text{ m}}$ .
Solving for the difference in radii: $|r_1 - r_2| = 10 \times 10^{-4} \text{ m} = 10^{-3} \text{ m} = 0.1 \text{ cm}$ .

Q122JEE Main 2017MCQ

A standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by $y(x,t)=0.5\text{sin}({5\pi }/4x)\text{cos}(200 \pi t)$ .What is the speed of the travelling wave moving in the positive x direction ?(x and t are in meter and second,respectively.)[Main 9 April 2017]

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📖 Explanation

The standing wave equation is given by $y(x,t) = 0.5 \sin\left(\frac{5\pi}{4}x\right) \cos(200\pi t)$ . Comparing this to the general form $y(x,t) = A \sin(kx) \cos(\omega t)$ , we identify the wave number $k = \frac{5\pi}{4} \text{ rad/m}$ and angular frequency $\omega = 200\pi \text{ rad/s}$ . The speed $v$ of the constituent traveling waves is determined by $v = \frac{\omega}{k}$ . Substituting the values, we get $v = \frac{200\pi}{5\pi/4} = \frac{200\pi \times 4}{5\pi} = 160 \text{ m/s}$ .

Q123JEE Main 2017MCQ

An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer?(speed of light $=3\times 10^8ms^{-1}$ )[Main 2017]

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📖 Explanation

Doppler effect in light(speed of observer is not very small compare to speed light) $f'=\sqrt{\frac{1+v/c}{1-v/c}}f_{\text{source}}=$ $\sqrt{\frac{1+1/2}{1-1/2}}$ (10 GHz)= 17.3 GHz

Q124JEE Main 2017MCQ

Two wires $W_1$ and $W_2$ have the same radius r and respective densities $\rho _1$ and $\rho _2$ such that $\rho _2=4\rho _1$ .They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in $W_1$ to $W_2$ is :[Main 8 April 2017]

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📖 Explanation

$n_1=n_2$ $T\to$ same $r\to$ same $I\to$ same $n=p/{2!}\sqrt{T/{\pi r^2d}}$ $n_1=n_2$ $p_1/\text{sqrtd}_1=p_2/\text{sqrtd}_2$ $p_1/p_2=1/2$

Q125JEE Main 2016MCQ

Two engines pass each other moving in opposite directions with uniform speed of 30 m/s. One of them is blowing a whistle of frequency 540 Hz. Calculate the frequency heard by driver of second engine before they pass each other. Speed of sound is 330 m/sec :[Main 9 Apr 2016]

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📖 Explanation

$f_0=(\frac{330+30}{330-30})540=648$ Hz

Explanation Figure 1

Q126JEE Main 2016MCQ

A toy-car, blowing its horn, is moving with a steady speed of 5 m/s, away from a wall.An observer, towards whom the toy car is moving, is able to hear 5 beats per second.If the velocity of sound in air is 340 m/s,the frequency of the horn of the toy car isclose to :[Main 10 Apr 2016]

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📖 Explanation

$F_\text{dir}=(340/{340-5})f$ $f_\text{ind}=(340/{340+5})f$ $f_\text{ind}-f_\text{dir}=5$ $\{(340/{340+5})(340/{340+5})\}f=5$ $340\{10/{(300-5)(340+5)}\}=5$ $f={340\times 5}/10$ $f=170Hz$

Explanation Figure 1

Q127JEE Main 2016MCQ

A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is : $(\text{take} g = 10 ms^{-2})$ [Main 2016]

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📖 Explanation

$T(x)={Mgx}/L$ $\Rightarrow v(x)=\text{sqrtT}/M=\sqrt{gx}$ $\Rightarrow \frac{dx}{dt}\sqrt{gx}\Rightarrow$ time taken $=2\sqrt2$ s.

Explanation Figure 1

Q128JEE Main 2016MCQ

A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now :[Main 2016]

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📖 Explanation

For open pipe in air, fundamental frequency: $f=V/{2l}$ For the pipe closed at one end (dipped in water), fundamental frequency: $f'=V/{4l}=V/{4\times l/2}=V/{2l}$ $\therefore f'=f$

Q129JEE Main 2015MCQ

A train is moving on a straight track with speed $20 ms^{-1}$ . It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is ( speed of sound $=320 ms^{-1}$ ) close to :[Main 2015]

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📖 Explanation

$f_1$ (train approaches) $=1000(320/{320-20})=1000(320/300)$ Hz $f_2$ (train approaches) $=1000(320/{320+20})=1000(320/300)$ Hz $\Delta f=({f_1-f_2}/f_1)\times 100%=(1-300/340)\times 100%=40/340\times 100%$ =11.7%≈12%

Q130JEE Main 2014MCQ

A pipe of length 85 cm is closed from one end.Find the number of possible natural oscillation of air column in the pipe whose frequencies lie below 1250 Hz.The velocity of sound in air is 340 m/s[Main 2014]

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📖 Explanation

In fundamental mode $\lambda /4=0.85$ $\lambda =4\times 0.85$ $\lambda =4\times 0.85$ $f=v/\lambda =340/{4\times 0.85}$ $=100$ Hz∴Possible frequencies =100 Hz,300 Hz,500 hz,700Hz,900 Hz 1100 Hz below 1250 Hz

Explanation Figure 1

Q131JEE Main 2013MCQ

A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %. What is the fundamental frequency of steel if density and elasticity of steel are $7.7 \times 10^3 {kg}/m^3$ and $2.2 \times 10^11N/m^2$ respectively?[Main 2013]

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📖 Explanation

Fundamental frequency $f =1/{2l}\sqrt{T/\mu }$ $=1/{2l}=\sqrt{T/{A\rho }}$ $=1/{2l}\sqrt{{stress}/\rho }=1/{2\times 1.5}\sqrt{\frac{2.2\times 10^11\times 10^{-2}}{7.7\times 10^3}}$

Q132JEE Main 2012MCQ

A cylindrical tube, open at both ends, has a fundamental frequency, $f$ , in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the aircolumn is now:

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📖 Explanation

The fundamental frequency of open tube $v_0=\;\frac{v}{2 l_0} \;\; . . .(i)$ That of closed pipe $v_c=\;\frac{v}{4 l_c} \;\; . . .(i i)$ According to the problem $l_c=\;\frac{l_0}{2}$ Thus $v_c=\;\frac{v}{l_0 / 2} \Rightarrow v_c \;\frac{v}{2 l} \;\; . . .$ (iii)From equations $(i)$ and $(i i i)$ $v_0=v_c$ Thus, $v_c=f \;\; (.$ as $v_0=f$ is given $)$

Q133JEE Main 2011MCQ

The transverse displacement $y(x, t)$ of a wave on a string is given by $y(x, t)=e^{- (a x^{2+}b t^{2+}2 {\sqrt{a b}} x t) }$ . This represents $a$ :

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📖 Explanation

Given wave equation is $\;y(x, t)=e (-a x^{2+}b t^{2+}2 {\sqrt{a b}} x t)$ $\;=e^{- \$ [({\sqrt{a x}})^{2+}({\sqrt{b t}})^{2+}2 {\sqrt{a}} x ⋅ {\sqrt{b}} t]$ };=e^{-({\sqrt{a}} x+{\sqrt{b}} t)^2};=e^{- (x+\sqrt{;\frac{b}{a}} t) ^2} $It is a function of type$ y=f(x+v t)\Rightarrow ;\text{ Speed of wave };=\sqrt{;\frac{b}{a}}$

Q134JEE Main 2010MCQ

The equation of a wave on a string of linear mass density $0.04 {\text{kg}} {\text{m}}^{-1}$ is given by $y=0.02(m) sin \; [2 \pi (\;\frac{t}{0.04(s)}-\;\frac{x}{0.50(m)}) ] .$ The tension in the string is

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📖 Explanation

$y=0.02(m) sin \; [2 \pi (\;\frac{t}{0.04(s)}) -\;\frac{x}{0.50(m)}]$ But $y=a sin \;(\omega t-k x)$ $\therefore \omega =\;\frac{2 \pi }{0.04} \Rightarrow v=\;\frac{1}{0.04}=25 {\text{Hz}}$ $k=\;\frac{2 \pi }{0.50} \Rightarrow \lambda =0.5 m$ $\therefore$ velocity, $v=v \lambda =25 \times 0.5 \frac{\text{m}}{\ s}=12.5 \frac{\text{m}}{\ s}$ Velocity on a string is given by $\;v=\sqrt{\;\frac{T}{\mu }}$ $\; \therefore T=v^2 \times \mu =(12.5)^2 \times 0.04=6.25 N$

Q135JEE Main 2009MCQ

Three sound waves of equal amplitudes have frequencies $(v-1), v,(v+1)$ . They superpose to give beats. The number of beats produced per second will be :

🚀 Solve in Practice Mode

📖 Explanation

Maximum number of beats $=(v+1)-(v-1)=2$

Q136JEE Main 2009MCQ

A motor cycle starts from rest and accelerates along a straight path at $2 \frac{\text{m}}{\ s}^2$ . At the starting point of the motor cycle there is a stationary electric siren. How far has the motor cycle gone when the driver hears the frequency of the siren at $94 \%$ of its value when the motor cycle was at rest? (Speed of sound $=330 {\text{ms}}^{-1}$ )

🚀 Solve in Practice Mode

📖 Explanation

$\;v_m^{2-}u^2=2 a s \Rightarrow v_m^2=2 \times 2 \times s$ $\; \therefore v_m=2 {\sqrt{s}}$ According to Doppler's effect $0.94 v=v [\;{330-2 {\sqrt{s}}}/{330}] \Rightarrow s=98.01 m$

Explanation Figure 1

Q137JEE Main 2008MCQ

While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of $18 {\text{cm}}$ during winter. Repeating the same experiment during summer, she measures the column length to be $x$ ${\text{cm}}$ for the second resonance. Then

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📖 Explanation

For first resonant length $v=\;\frac{v}{4 ℓ_1}=\;\frac{v}{4 \times 18}$ (in winter)For second resonant length $\;v^{'}=\;\frac{3 v^{'}}{4 ℓ_2}=\;\frac{3 v^{'}}{4 x} \;\text{ (\in summer) }\;$ $\; \therefore \;\frac{v}{4 \times 18}=\;\frac{3 v^{'}}{4 \times x}$ $\; \therefore x=3 \times 18 \times \;\frac{v^{'}}{v}$ $\; \therefore x=54 \times \;\frac{v^{'}}{v} {\text{cm}}$ $v^{'}>v$ because velocity of light is greater in summer as compared to winter $\;(v \propto {\sqrt{T}})$ $\; \therefore x>54 {\text{cm}}$

Q138JEE Main 2008MCQ

A wave travelling along the $x$ -axis is described by the equation $y(x, t)=0.005$ $\cos (\alpha x-\beta t)$ . If the wavelength and the time period of the wave are $0.08 {\text{m}}$ and $2.0 {\ s}$ , respectively, then $\alpha$ and $\beta$ in appropriate units are

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📖 Explanation

$y(x, t)=0.005 \cos (\alpha x-\beta t) \;\text{ (Given) }\;$ Comparing it with the standard equation of wave $y(x, t)=a \cos (k x-\omega t)$ we get $k=\alpha \;\;$ and $\;\; \omega =\beta$ $\therefore \;\frac{2 \pi }{\gamma }=\alpha \;\;$ and $\;\; \;\frac{2 \pi }{T}=\beta$ $\therefore \alpha =\;\frac{2 \pi }{0.08}=25 \pi \;\;$ and $\;\; \beta =\;\frac{2 \pi }{2}=\pi$

Q139JEE Main 2007MCQ

A sound absorber attenuates the sound level by $20 {\text{dB}}$ . The intensity decreases by a factor of

🚀 Solve in Practice Mode

📖 Explanation

We have, $L_1=10 \log (\;{ {\text{I}}_1}/{ {\text{I}}_0})$ ; $L_2=10 \log (\;{ {\text{I}}_2}/{ {\text{I}}_0})$ $\therefore L_1-L_2=10 \log (\;{ {\text{I}}_1}/{ {\text{I}}_0}) -10 \log (\;{ {\text{I}}_2}/{ {\text{I}}_0})$ or, $\Delta L=10 \log (\;\frac{I_1}{I_0} \times \;{ {\text{I}}_0}/{ {\text{I}}_2})$ or, $\Delta L=10 \log (\;{ {\text{I}}_1}/{ {\text{I}}_2})$ or, $20=10 \log (\;\frac{I_1}{I_2})$ or, $2=\log (\;\frac{I_1}{I_2})$ or, $\;\frac{I_1}{I_2}=10^2$ or, ${\text{I}}_2=\;{ {\text{I}}_1}/{100}$ . $\Rightarrow$ Intensity decreases by a factor 100 .

Q140JEE Main 2006MCQ

A string is stretched between fixed points separated by $75.0 {\text{cm}}$ . It is observed to have resonant frequencies of $420 {\text{Hz}}$ and $315 {\text{Hz}}$ . There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

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📖 Explanation

Given $\;\frac{n v}{2 ℓ}=315$ and $(n+1) \;\frac{v}{2 ℓ}=420$ $\Rightarrow \;\frac{n+1}{n}=\;\frac{420}{315} \Rightarrow n=3$ Hence $3 \times \;\frac{v}{2 ℓ}=315 \Rightarrow \;\frac{v}{2 ℓ}=105 {\text{Hz}}$ Lowest resonant frequency is when $n=1$ Therefore lowest resonant frequency $=105 {\text{Hz}}$ .

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