Waves Physics: Key Concepts, Formulas & Previous Year JEE Questions with Solutions

Waves Previous Year Questions With Solutions

1.0Introduction

A wave is a disturbance that transfers energy from one point to another without moving matter. Waves can travel through different mediums like air, water, or solids, and some can even move through a vacuum, such as light. There are two main types: mechanical waves, which require a medium (like sound or water waves), and electromagnetic waves, which do not (like light and radio waves). Waves can be transverse, where particles move perpendicular to the wave, or longitudinal, where particles move parallel. Key characteristics include amplitude, wavelength, frequency, and speed.

2.0Key Concepts to Remember

Wave Function

The wave function represents the displacement of particles in a medium. For a wave on a string, it’s the displacement; for sound, it’s pressure or density fluctuations.

General wave equation:
y(x,t)=fx∓vt where v is the wave speed.

When two waves meet, their displacements combine:

Constructive interference: Amplitudes add.
Destructive interference: Amplitudes subtract.
Resultant wave: y=y1+y2

Standing Waves

Standing waves, characterized by nodes and antinodes, result from the interference of identical waves moving in opposite directions.

Equation: $y (x, t) = (2 A cos k x) sin ω t$

Sound Waves

Sound waves are longitudinal waves requiring a medium (solid, liquid, or gas) to propagate. It consists of compressions and rarefactions.

Intensity of Sound Waves

Intensity is the energy transmitted per unit time and per unit area, and it is proportional to the square of the pressure amplitude.

Pitch and Frequency

Pitch: A psychological perception related to frequency.

Frequency: The number of oscillations per second, measured in Hz.

Loudness and Intensity

Loudness is a subjective perception, while intensity is an objective measure of sound energy. Loudness increases logarithmically with intensity.

Interference of Sound Waves

Constructive interference: Occurs when waves are in phase.

Destructive interference: Occurs when waves are out of phase.

Beats

Slightly different frequency waves interfering produce beats: periodic intensity variations.

Doppler Effect

When there is relative motion between a sound or light wave source and an observer along the line connecting them, the frequency observed differs from the source's frequency. This phenomenon is known as the Doppler Effect.

3.0Important Formulas

General Wave Equation	$y (x, t) = f (x \pm v t) where v is the wave speed$
Phase velocity	$v = - \frac{Coefficient of t}{Coefficient of x}$
Wave on a String	$y (x, t) = A sin (k x - ω t)$
Wave speed	$v = \frac{T}{μ}$
Power Transmitted by a Sine Wave	$P_{avg} = 2 π^{2} μv A^{2} f^{2} \Rightarrow P_{avg} = \frac{1}{2} μv A^{2} ω^{2}$
For reflection and transmission	$A_{r} = (\frac{2 μ _{1}}{μ _{1} + μ _{2}}) A_{i} and A_{t} = (\frac{μ _{1} - μ _{2}}{μ _{1} + μ _{2}}) A_{i}$
Standing Waves	$y (x, t) = (2 A cos k x) sin ω t$
Laws of Transverse Vibrations of a String	Length Law: $f \propto \frac{1}{l}$ Tension Law: $f \propto T$ Mass Law: $f \propto \frac{1}{μ}$
Beat Frequency	$∣ f_{1} - f_{2} ∣$
Closed Organ Pipe	$f_{0} = \frac{v}{λ _{0}} = \frac{v}{4 l} (λ_{0} = 4 l) n^{th} o v er t o n e f_{n} = (2 n + 1) f_{0}$
Open Organ Pipe	$f_{0} = \frac{v}{λ _{0}} = \frac{v}{2 l} (λ_{0} = 2 l) n^{th} o v er t o n e f_{n} = (n + 1) f_{0}$
End Correction	Closed pipe: l+0.6r Open pipe: l+1.2r
Doppler Effect	$f^{'} = f (\frac{v \pm v _{o}}{v \pm v _{s}})$

4.0Past Year Questions with Solutions on Waves: JEE (Mains)

Q-1.A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of length 350 cm, both vibrating in fundamental mode. The velocity of sound is ________ m/s.

Solution: Ans(294)

$f_{c} = \frac{v}{4 l _{1}}, f_{o} = \frac{v}{2 l _{2}} f_{c} - f_{o} = 7 \frac{v}{4 \times 150} - \frac{v}{2 \times 350} = 7 \frac{v}{600 cm} - \frac{v}{700 cm} = 7 \frac{v}{6 m} - \frac{v}{7 m} = 7$

Q-2.In a closed organ pipe, the frequency of the fundamental note is 30 Hz. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 Hz. If the organ pipe has a cross-sectional area of 2 cm², the amount of water poured in the organ tube is _______ g. (Take speed of sound in air is 330 m/s)

Solution: Ans(400)

$\frac{v}{4 l _{1}} = 30 \Rightarrow l_{1} = \frac{11}{4} m \frac{v}{4 l _{2}} = 110 \Rightarrow l_{1} = \frac{3}{4} m Δ l = 2 m Change in volume = A Δ l M = 400 g (∵ ρ = 1 g / c m^{3})$

Q-3.The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe will be :

1) 60 cm 2) 45 cm 3) 30 cm 4) 15 cm

Solution: Ans(4)

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60 cm

$Case-1 \frac{λ}{4} = L_{1} v = f λ v = f_{1} (4 L_{1}) f_{1} = \frac{v}{4 L _{1}} Case-2 2 (\frac{λ}{2}) = λ f_{2} = \frac{v}{L _{2}} f_{1} = f_{2} \frac{v}{4 L _{1}} = \frac{v}{L _{2}} \Rightarrow L_{2} = 4 L_{1} 60 = 4 \times L_{1} L_{1} = 15 c m$

Q-4. Two waves of intensity ratio 1 : 9 cross each other at a point. The resultant intensities at the point, when (a) Waves are incoherent is I₁(b) Waves are coherent is I₂ and differ in phase by 60°. If $\frac{I _{1}}{I _{2}} = \frac{10}{x}$ then x = _____.

Solution: Ans(13)

$I_{2} = I_{A} + I_{B} + 2 I_{A} I_{B} cos 6 0^{\circ} \Rightarrow I_{2} = I_{0} + 9 I_{0} + 2 9 I_{0} \cdot I_{0} \cdot \frac{1}{2} = 13 I_{0} \frac{I _{1}}{I _{2}} = \frac{10}{13}$

Q-5. The speed of sound in oxygen at S.T.P. will be approximately:

(Given,R=8.3JK-1,=1.4 )

1) 315 m/s 2) 333 m/s 3) 341 m/s 4) 325 m/s

Solution: Ans(1)

$v = \frac{γ RT}{M} = \frac{1.4 \times 8.3 \times 273}{32 \times 1 0 ^{- 3}} = 314.8541 \approx 315 m/s$

Q-6.A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 6 N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is _______ Hz.

Solution: Ans(6)

$f = \frac{1}{2 L} \frac{T}{μ} f_{1} = \frac{1}{2} \frac{6}{μ} ...... (1) f_{2} = \frac{1}{2} \frac{54}{μ} ........ (2) \frac{f _{1}}{f _{2}} = \frac{1}{3} \Rightarrow f_{2} = 3 f_{1} f_{2} - f_{1} = 12 3 f_{1} - f_{1} = 12 \Rightarrow f_{1} = 6 Hz$

Q-7. Two open organ pipes of length 60 cm and 90 cm resonate at 6^th and 5^th harmonics respectively. The difference of frequencies for the given modes is _____ Hz.(Velocity of sound in air = 333 m/s)

Solution:Ans(740)

The difference in frequency in open organ pipe,

$f = \frac{n v}{2 L} Δ f = \frac{6 v}{2 \times 0.6} - \frac{5 v}{2 \times 0.9} v = 333 m / s Δ f = 740 Hz$

Q-8.A closed and an open organ pipe have the same lengths. If the ratio of frequencies of their seventh overtones is a-1a (\frac {a-1}{a}) then the value of a is _____.

Solution:Ans(16)

$For closed organ pipe f_{c} = (2 n + 1) \frac{v}{4 l} = \frac{15 v}{4 l} For open organ pipe: f_{o} = (n + 1) \frac{v}{2 l} = \frac{8 v}{2 l} \frac{f _{c}}{f _{o}} = \frac{15}{16} = \frac{a - 1}{a} \Rightarrow a = 16$

Q-9. A sonometer wire of resonating length 90 cm has a fundamental frequency of 400 Hz when kept under some tension. The resonating length of the wire with fundamental frequency of 600 Hz under the same tension _______ cm.

Solution: Ans(60)

$f_{0} = 400 Hz, v = \frac{T}{μ} = Constant \frac{λ}{2} = L, v = f_{0} λ \Rightarrow \frac{v}{2 f _{0}} = L \Rightarrow v = 2 L f_{0} L^{'} = \frac{v}{2 f ^{'}} = \frac{2 L f _{0}}{2 f ^{'}} = \frac{L f _{0}}{f ^{'}} = \frac{90 \times 400}{600} = 60 cm$

Q-10.A plane progressive wave is given by y = 2 cos 2 (330 t – x) m. The frequency of the wave is :

1) 165 Hz 2) 330 Hz 3) 660 Hz 4 )340 Hz

Solution: Ans(2)

$y = 2 cos 2 π (330 t - x) m y = A cos (ω t - k x) ω = 2 π \times 330 2 π f = 2 π \times 330 \Rightarrow f = 330 Hz$

Q-11.Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm²) is 90 ms^–1. If the Young's modulus of wire is 16 × 10¹¹ Nm^–2, the extension of wire over its natural length is :

1) 0.02 mm 2) 0.04 mm 3) 0.03 mm 4) 0.01 mm

Solution: Ans(3)

$v = \frac{T}{μ} 90 = \frac{Y A Δ l}{\frac{m}{l}} = \frac{16 \times 1 0 ^{11} \times 1 0 ^{- 6} \times Δ l}{6 \times 1 0 ^{- 3}} = \frac{8100 \times 3}{8} \times 1 0^{- 8} = Δ l$

Q-12.A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is n₀ = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to :

$(1) \frac{1}{8} m/s$

$(2) \frac{1}{2} m/s$

$(3) 1 m/s$

$(4) \frac{1}{4} m/s$

Solution: Ans(4)

A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound)

$f_{1} = (\frac{c}{c - v}) f_{0}, f_{2} = (\frac{c}{c + v}) f_{0} B e a tF re q u e n cy (Δ f) = f_{1} - f_{2} = f_{0} [(\frac{1}{1 - \frac{v}{c}}) - (\frac{1}{1 + \frac{v}{c}})] On solving \frac{Δ f}{f _{0}} = \frac{2 v}{c} \Rightarrow v = \frac{Δ f}{2 f _{0}} \times c = \frac{2 \times 350}{2 \times 1400} = \frac{1}{4} m/s$

Q-13.A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is ________ Hz.

Solution: Ans. (106.00 to 107.20)

$\frac{v _{gas}}{v _{air}} = \frac{ρ _{air}}{ρ _{gas}} \Rightarrow \frac{v _{gas}}{300} = \frac{1}{2} v_{gas} = \frac{300}{2} v_{gas} = 1502 n_{2} - n_{1} = \frac{v _{gas}}{2 l} = \frac{150 2}{2 ( 1 )} = 752 Δ n = 106.06 Hz$

Q-14.A transverse wave travels on a taut steel wire with a velocity of v when tension in it is 2.06 × 10⁴ N. When the tension is changed to T, the velocity changed to v/2. The value of T is close to :

1) $10.2 \times 1 0^{2} N$

2) $5.15 \times 1 0^{3} N$

3) $2.50 \times 1 0^{4} N$

4) $30.5 \times 1 0^{4} N$

Solution: Ans(2)

$Velocity of transverse wave V \propto T V \to \frac{V}{2} \Rightarrow T \to T^{'} = \frac{T}{4} T^{'} = \frac{2.06 \times 1 0 ^{4}}{4} = 5.15 \times 1 0^{3} N$

Q-15.A wire of length L and mass per unit length 6.0 × 10^–3 kgm^–1 is put under tension of 540 N. Two consecutive frequencies that it resonates at are : 420 Hz and 490 Hz. Then L in meters is :

1) 8.1 m 2) 5.1 m 3) 1.1 m 4) 2.1 m

Solution: Ans(2.1 m)

$\frac{n v}{2 l} = 420 \frac{( n + 1 ) v}{2 l} = 490 \Rightarrow \frac{v}{2 l} = 70 l = \frac{v}{140} = \frac{1}{140} \frac{540}{6 \times 1 0 ^{- 3}} = \frac{1}{140} 90 \times 1 0^{3} l = \frac{300}{140} = 2.142 m$

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