EAPCET Physics · Intermediate second year

Waves

Waves covers wave motion, the speed of waves in strings and gases, standing waves in strings and pipes, beats and the Doppler effect.

Subject
Physics
Taught in
Intermediate second year
Physics in the paper
40 of 160 questions

Key formulas and results

  • v=fλv = f\lambda
  • Speed on a string: v=T/μv = \sqrt{T/\mu}
  • Speed of sound (Laplace): v=γP/ρv = \sqrt{\gamma P/\rho}
  • Open pipe: all harmonics, fn=nv2Lf_{n} = \frac{nv}{2L}; closed pipe: odd harmonics, f=(2n−1)v4Lf = \frac{(2n - 1)v}{4L}
  • Beat frequency =∣f1−f2∣= |f_{1} - f_{2}|
  • Doppler: f′=f v±v0v∓vsf' = f\,\dfrac{v \pm v_{0}}{v \mp v_{s}}
  • Resonance tube (first two resonating lengths l1l_{1} and l2l_{2}): v=2f(l2−l1)v = 2f(l_{2} - l_{1})
  • Progressive wave: y=Asin⁡(ωt−kx+ϕ)y = A\sin(\omega t - kx + \phi), with k=2πλk = \frac{2\pi}{\lambda} and phase difference Δϕ=2πλΔx\Delta\phi = \frac{2\pi}{\lambda}\Delta x
  • Particle velocity in a wave: vparticle=−v×(slope ∂y∂x)v_{\text{particle}} = -v \times \left(\text{slope } \frac{\partial y}{\partial x}\right)
  • Energy density u=12ρA2ω2u = \tfrac{1}{2}\rho A^{2}\omega^{2}; intensity I=12ρA2ω2vI = \tfrac{1}{2}\rho A^{2}\omega^{2}v, so I∝A2I \propto A^{2}
  • Intensity from a point source: I=P4πr2I = \frac{P}{4\pi r^{2}}, so I∝1r2I \propto \frac{1}{r^{2}}
  • Superposition: resultant amplitude A=A12+A22+2A1A2cos⁡ϕA = \sqrt{A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2}\cos\phi}; Imax⁡Imin⁡=(I1+I2)2(I1−I2)2\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_{1}} + \sqrt{I_{2}})^{2}}{(\sqrt{I_{1}} - \sqrt{I_{2}})^{2}}
  • Sound intensity level: β=10log⁡10(I/I0) dB\beta = 10\log_{10}(I/I_{0})\ \text{dB}, and β2−β1=10log⁡10(I2/I1)\beta_{2} - \beta_{1} = 10\log_{10}(I_{2}/I_{1})
  • Resonance tube: λ=2(l2−l1)\lambda = 2(l_{2} - l_{1}), v=2f(l2−l1)v = 2f(l_{2} - l_{1}), end correction e=l2−3l12≈0.3de = \frac{l_{2} - 3l_{1}}{2} \approx 0.3d
  • Speed of sound in a gas: v=γRT/Mv = \sqrt{\gamma RT/M}

What this chapter covers

  • Transverse and longitudinal waves
  • Speed of waves
  • Superposition and standing waves
  • Strings and organ pipes
  • Beats
  • Doppler effect

How it is usually asked

  • Harmonics in strings and pipes
  • Beats
  • Doppler effect
  • Reading a wave equation

Understanding Waves

A wave carries energy from place to place without carrying the medium along. In transverse waves the particles move perpendicular to the direction of travel; in longitudinal waves such as sound they move along it. Wave speed is set by the medium: on a string it is √(T/μ), and in a gas it is √(γP/ρ). Frequency is fixed by the source, so v = fλ decides the wavelength.

When two waves overlap, their displacements add. Two equal waves travelling in opposite directions form a standing wave with fixed nodes and antinodes. A string fixed at both ends and an open pipe support all harmonics, while a pipe closed at one end supports only odd harmonics. Two slightly different frequencies produce beats, and relative motion of source and observer shifts the heard frequency (the Doppler effect).

EAPCET usually tests harmonics and overtones in strings and pipes, beats, Doppler shifts and reading a wave equation such as y = A sin(kx − ωt), where the speed is ω/k, the wavelength 2π/k and the frequency ω/2π. In beat problems, loading a tuning fork with wax lowers its frequency, and the resulting change in beat frequency tells you which of the two possible values was correct.

Worked example

EAPCET-style question

A pipe closed at one end has a length of 0.85 m. If the speed of sound in air is 340 m/s (ignore end correction), the frequency of its first overtone is

  1. (1)300 Hz
  2. (2)200 Hz
  3. (3)100 Hz
  4. (4)400 Hz

Solution

  1. Fundamental of a closed pipe: f1=v4L=3404×0.85=3403.4=100 Hzf_{1} = \frac{v}{4L} = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100\ \text{Hz}.
  2. A closed pipe has only odd harmonics: f1,3f1,5f1,…f_{1}, 3f_{1}, 5f_{1}, \ldots
  3. The first overtone is the next allowed mode, the third harmonic.
  4. First overtone =3×100=300 Hz= 3 \times 100 = 300\ \text{Hz}.

Answer: 300 Hz

Shortcuts and results to remember

  • Closed pipe: nnth overtone is the (2n+1)(2n + 1)th harmonic; open pipe: nnth overtone is the (n+1)(n + 1)th harmonic.
  • A wave reflected from a fixed end is inverted (phase change π\pi); from a free end it is not inverted.
  • Filing the prongs of a tuning fork raises its frequency, while loading them with wax lowers it.
  • An open pipe of length ll has the same fundamental frequency as a closed pipe of length l/2l/2.
  • Audible range is 20 Hz to 20 kHz; below is infrasonic and above is ultrasonic.

More practice questions

Practice question 1

A stationary wave on a string of linear density 0.3 kg/m is y=20sin⁡(πx/2)cos⁡(20πt)y = 20\sin(\pi x/2)\cos(20\pi t), with xx in metres. The tension in the string is

  1. (1)120 N
  2. (2)240 N
  3. (3)480 N
  4. (4)960 N

Solution

  1. ω=20π\omega = 20\pi, so f=10 Hzf = 10\ \text{Hz}
  2. k=π/2k = \pi/2, so λ=2π/k=4 m\lambda = 2\pi/k = 4\ \text{m}
  3. Wave speed v=fλ=40 m/sv = f\lambda = 40\ \text{m/s}
  4. T=μv2=0.3×1600=480 NT = \mu v^{2} = 0.3 \times 1600 = 480\ \text{N}

Answer: 480 N

Practice question 2

The speed of sound in nitrogen (M=28 g/molM = 28\ \text{g/mol}, γ=7/5\gamma = 7/5) at 27 °C is about (R=8.314 J mol−1 K−1R = 8.314\ \text{J mol}^{-1}\,\text{K}^{-1})

  1. (1)280 m/s
  2. (2)330 m/s
  3. (3)353 m/s
  4. (4)410 m/s

Solution

  1. v=γRT/Mv = \sqrt{\gamma RT/M} with T=300 KT = 300\ \text{K} and M=0.028 kg/molM = 0.028\ \text{kg/mol}
  2. γRT=1.4×8.314×300≈3492\gamma RT = 1.4 \times 8.314 \times 300 \approx 3492
  3. γRT/M≈3492/0.028≈1.247×105\gamma RT/M \approx 3492/0.028 \approx 1.247 \times 10^{5}
  4. v≈353 m/sv \approx 353\ \text{m/s}

Answer: 353 m/s

Common mistakes to avoid

  • Treating the first overtone of a closed pipe as the second harmonic — closed pipes have only odd harmonics, so it is the third harmonic.
  • Using v/2L for a closed pipe — that is the open-pipe fundamental; a closed pipe has v/4L.
  • Putting the volume density of the wire in v = √(T/μ) — μ is mass per unit length in kg/m.
  • Choosing the Doppler signs by memory without a sanity check — when source and observer approach, the heard frequency must rise.

How to prepare it

In the Doppler formula, the signs make the frequency rise when source and observer approach — use that to check.

Once the formulas feel familiar, test them the way the exam does: practise Waves questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Waves chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Waves part of the EAPCET Physics syllabus?
Yes. Waves is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Waves for EAPCET?
In the Doppler formula, the signs make the frequency rise when source and observer approach — use that to check.
How many Physics questions are there in EAPCET?
40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Waves?
Treating the first overtone of a closed pipe as the second harmonic — closed pipes have only odd harmonics, so it is the third harmonic. Using v/2L for a closed pipe — that is the open-pipe fundamental; a closed pipe has v/4L. Putting the volume density of the wire in v = √(T/μ) — μ is mass per unit length in kg/m. Choosing the Doppler signs by memory without a sanity check — when source and observer approach, the heard frequency must rise.