Questions: Waves

Q1 What is a wave? Distinguish between transverse and longitudinal waves.

Answer (English): A wave is a disturbance that travels through a medium, transferring energy and momentum from one point to another without permanently displacing the medium. In a transverse wave the particles oscillate perpendicular to the direction of propagation, as in a wave on a stretched string, and the wave can be polarised. In a longitudinal wave the particles oscillate parallel to the direction of propagation, as in sound in air, where the disturbance is a series of compressions and rarefactions. Transverse waves can travel through solids and on liquid surfaces, while longitudinal waves travel through solids, liquids and gases.

Hinglish: Wave disturbance hai jo energy aur momentum aage le jaati hai, medium ko permanently displace nahi karti. Transverse mein particles propagation ke perpendicular oscillate karte hain (string), longitudinal mein parallel (sound, compression-rarefaction).

Q2 A wave is described by y = 4 sin(20t - 0.5x), with x, y in cm and t in s. Find the amplitude, frequency, wavelength and speed.

Answer (English): Comparing with y = A sin(omega t - kx), the amplitude is 4 centimetres, the angular frequency is 20 radians per second and the wave number is 0.5 radians per centimetre. The frequency is 20 divided by 2 pi, about 3.2 hertz, and the wavelength is 2 pi divided by 0.5, about 12.6 centimetres. The speed is omega divided by k, which is 40 centimetres per second, or 0.40 metre per second. Since the x and t terms in the argument have opposite signs, the wave travels along the positive x direction.

Hinglish: Standard form se A = 4 cm, ω = 20 rad/s, k = 0.5 rad/cm. Thus f = ω/2π ≈ 3.2 Hz, λ = 2π/k ≈ 12.6 cm, v = ω/k = 40 cm/s. Argument mein x aur t ke signs opposite hain, isliye wave +x direction mein chalti hai.

Q3 A string of length 2.0 m and mass 20 g is stretched with a tension of 80 N. Find the speed of a transverse wave and the fundamental frequency.

Answer (English): The linear mass density is the mass divided by the length, 0.020 kilogram divided by 2.0 metres, which is 0.010 kilogram per metre. The wave speed is the square root of the tension divided by the linear density, the square root of 80 divided by 0.010, which is the square root of 8000, about 89.4 metres per second. The fundamental frequency of a string fixed at both ends is the speed divided by twice the length, which is 89.4 divided by 4.0, about 22.4 hertz.

Hinglish: μ = 0.020/2.0 = 0.010 kg/m. v = √(T/μ) = √(80/0.010) = √8000 ≈ 89.4 m/s. Fundamental f1 = v/2L = 89.4/4.0 ≈ 22.4 Hz.

Q4 Show that the speed of sound in a gas is independent of pressure.

Answer (English): Start from the Laplace formula, the speed equals the square root of gamma P divided by rho. For an ideal gas, the density equals the molar mass times pressure divided by R T, so P divided by rho equals R T divided by the molar mass. Substituting this gives the speed as the square root of gamma R T divided by the molar mass, which contains only the temperature, the gas constant and the molar mass. Pressure has cancelled, so at a fixed temperature the speed does not change when the pressure changes. If the pressure rises at constant temperature, the density rises in the same proportion, and the ratio stays the same.

Hinglish: v = √(γP/ρ) se shuru karo. Ideal gas ke liye P/ρ = RT/M, isliye v = √(γRT/M). Ismein pressure nahi bacha, sirf temperature hai. Pressure badhne par density bhi utni hi badh jaati hai, ratio same rehta hai.

Q5 Derive the frequency of the fundamental mode of a string fixed at both ends and give the frequency of the nth harmonic.

Answer (English): Take a string of length L clamped at both ends, so each end is forced to stay still. If the string vibrates in one loop, the loop length is half a wavelength, so L equals half of lambda and the wavelength is twice L. The fundamental frequency is the wave speed divided by the wavelength, which is v divided by 2L. For n loops the string holds n times half a wavelength, so L equals n lambda over 2 and the nth harmonic frequency is n times v divided by 2L. All integer multiples of the fundamental are allowed.

Hinglish: Dono ends par node hone chahiye, isliye L = nλ/2. n = 1 ke liye λ = 2L aur f1 = v/2L; nth harmonic f_n = nv/2L. Isliye string par saare harmonics milte hain.

Q6 Why does a closed organ pipe produce only odd harmonics? Give the fundamental frequency.

Answer (English): In a closed pipe one end is a displacement node and the other is a displacement antinode. The shortest standing wave that fits has a quarter wavelength in the pipe, so the fundamental wavelength is four times L and the fundamental frequency is v divided by 4L. The next allowed pattern has three quarter wavelengths, then five, and so on, because the closed end must remain a node and the open end an antinode. The general wavelength is four L divided by the odd number, so the frequencies are v over 4L, 3v over 4L, 5v over 4L. Only odd multiples appear, and the even harmonics are absent.

Hinglish: Closed end node, open end antinode. Sabse chhoti standing wave λ/4 ki hoti hai, isliye λ1 = 4L aur f1 = v/4L. Baad mein 3λ/4, 5λ/4 fit hote hain, isliye sirf odd harmonics milte hain; even harmonics absent hain.

Q7 Two identical waves of amplitude A travelling in opposite directions superpose. Find the resultant and the node spacing.

Answer (English): Adding the two waves gives a standing wave of the form 2A sin(kx) cos(omega t). The amplitude of vibration at position x is 2A sin(kx). The nodes are the points where the sine is zero, which happens at kx equal to n pi, that is at x equal to n lambda over 2. Adjacent nodes are therefore separated by half a wavelength, and the antinodes lie halfway between them, at odd multiples of lambda over 4. The maximum amplitude at an antinode is twice the amplitude of each travelling wave.

Hinglish: Opposite travelling waves add hokar y = 2A sin(kx) cos(ωt) dete hain. Nodes par sin(kx) = 0, yaani x = nλ/2, isliye node spacing = λ/2 aur antinode node se λ/4 door hota hai.

Q8 A pipe closed at one end has length 0.85 m. Taking the speed of sound as 340 m/s and ignoring end correction, find the first three natural frequencies.

Answer (English): A closed pipe gives frequencies v over 4L times odd numbers. The fundamental is 340 divided by 3.4, which is 100 hertz, because four times the length is 3.4 metres. The next allowed frequencies are the third harmonic, 300 hertz, and the fifth harmonic, 500 hertz. The even multiples, 200 and 400 hertz, are not natural frequencies of this pipe. Including the end correction would lower each frequency slightly, because the effective length would be a little more than 0.85 metre.

Hinglish: Closed pipe: f_n = (2n-1)v/4L. Yahan 4L = 3.4 m, isliye f1 = 100 Hz, f3 = 300 Hz, f5 = 500 Hz. Even harmonics (200, 400 Hz) nahi honge; end correction se actual frequencies thodi kam hongi.

Q9 Two tuning forks of frequencies 384 Hz and 390 Hz are sounded together, and a third fork of unknown frequency produces 3 beats per second with the first fork and 9 beats per second with the second. Find the unknown frequency.

Answer (English): A beat count of 3 with the 384 hertz fork means the unknown frequency is either 387 or 381 hertz. A beat count of 9 with the 390 hertz fork means it is either 399 or 381 hertz. The only value that satisfies both readings is 381 hertz, so the unknown frequency is 381 hertz. The two candidates are tested one by one, and the one that fits both observations is kept.

Hinglish: 384 se 3 beats ka matlab 387 ya 381 Hz; 390 se 9 beats ka matlab 399 ya 381 Hz. Dono ko satisfy karne wali value sirf 381 Hz hai, isliye unknown frequency 381 Hz hai.

Q10 A source of frequency 500 Hz moves towards a stationary observer at 34 m/s. If the speed of sound is 340 m/s, find the observed frequency. What happens after it passes the observer?

Answer (English): The source approaches the observer, so the denominator uses the minus sign with the source speed: the observed frequency is 500 times 340 divided by 340 minus 34, which is 500 times 340 over 306, about 556 hertz. After the source passes the observer it recedes, so the denominator uses the plus sign and the observed frequency becomes 500 times 340 over 374, about 455 hertz. The pitch drops as the source goes past, which is the familiar change heard from a passing siren.

Hinglish: Approach mein f' = f v/(v - v_s) = 500 × 340/306 ≈ 556 Hz. Pass hone ke baad recession mein f' = 500 × 340/374 ≈ 455 Hz. Isliye siren ki pitch pass hote waqt girti hai.

That completes the chapter questions. Try the quick quiz below, then move on to the previous year questions.