Waves Class 11 Physics NCERT Solutions

Step-by-step solution

1. 1Linear mass density μ = m/L = 2.50/20.0 = 0.125 kg m⁻¹.
2. 2Wave speed v = √(T/μ) = √(200/0.125) = √1600 = 40 m s⁻¹.
3. 3Time t = L/v = 20.0/40 = 0.5 s.

Step-by-step solution

1. 1Fall time for the stone: t₁ = √(2h/g) = √(2 × 300/9.8) = √61.2 = 7.82 s.
2. 2Time for the sound of the splash to reach the top: t₂ = h/v = 300/340 = 0.882 s.
3. 3Total time = t₁ + t₂ = 7.82 + 0.882 ≈ 8.7 s.

Step-by-step solution

1. 1μ = m/L = 2.10/12.0 = 0.175 kg m⁻¹, and we require v = 343 m s⁻¹.
2. 2v = √(T/μ) gives T = μv².
3. 3T = 0.175 × 343² = 0.175 × 117649 ≈ 2.06 × 10⁴ N.

Step-by-step solution

1. 1(a) For an ideal gas ρ = PM/RT, so v = √(γP/ρ) = √(γRT/M). The pressure cancels — v is independent of P.
2. 2(b) From v = √(γRT/M), the speed is proportional to √T, so it increases with temperature.
3. 3(c) Water vapour (M = 18) is lighter than dry air (mix, M ≈ 29); moist air therefore has a smaller average M and a larger ρ⁻¹ effect, so v increases with humidity.

Step-by-step solution

1. 1The converse holds only for bounded functions — the function must be a finite, pulse-like form of (x ∓ vt).
2. 2(a) (x − vt)² has the right combination but grows without bound as x, t grow — it is not a physically realisable travelling pulse.
3. 3(b) log[(x + vt)/x0] is of the form f(x + vt) and bounded over a finite range — represents a wave travelling along −x.
4. 4(c) 1/(x + vt) is of the form f(x + vt) — represents a wave travelling along −x.

Step-by-step solution

1. 1The frequency of the source is f = 10³ kHz = 10⁶ Hz, unchanged on reflection or transmission.
2. 2(a) Reflected (in air): λ_air = v_air/f = 340/10⁶ = 3.4 × 10⁻⁴ m.
3. 3(b) Transmitted (in water): λ_water = v_water/f = 1486/10⁶ = 1.5 × 10⁻³ m.

Step-by-step solution

1. 1v = 1.7 km s⁻¹ = 1700 m s⁻¹, f = 4.2 MHz = 4.2 × 10⁶ Hz.
2. 2λ = v/f = 1700/(4.2 × 10⁶) = 4.0 × 10⁻⁴ m.

Step-by-step solution

1. 1(a) x and t appear together in (36t + 0.018x) — a travelling wave. Comparing with y = A sin(ωt + kx + φ), the +kx term with +ωt gives ωt + kx = constant, so x = −(ω/k)t: the wave travels towards the left with speed v = ω/k = 36/0.018 = 2000 cm s⁻¹ = 20 m s⁻¹.
2. 2(b) A = 3.0 cm; f = ω/2π = 36/2π = 5.73 Hz.
3. 3(c) At x = 0, the phase is 36t + π/4, so the initial phase at the origin is π/4.
4. 4(d) Successive crests are one wavelength apart: λ = 2π/k = 2π/0.018 = 349 cm ≈ 3.5 m.

Step-by-step solution

1. 1For fixed x the wave gives y = 3.0 sin(36t + φ₀) with a constant φ₀ = 0.018x + π/4.
2. 2At x = 0, 2, 4 cm the graphs are pure sine curves of the same amplitude (3 cm) and the same frequency (5.73 Hz).
3. 3Only the constant phases differ: φ₀ = π/4, 0.036 + π/4, 0.072 + π/4 — i.e. the graphs are shifted along the time axis.

Step-by-step solution

1. 1The spatial coefficient is 2π × 0.0080 cm⁻¹, so λ = 1/0.0080 = 125 cm = 1.25 m.
2. 2Phase difference Δφ = (2π/λ) Δx.
3. 3(a) Δx = 400 cm: Δφ = 2π × 3.2 = 6.4π ≈ 20 rad.
4. 4(b) Δx = 50 cm: Δφ = 2π × 0.4 = 0.8π ≈ 2.5 rad.
5. 5(c) Δx = λ/2: Δφ = π rad.
6. 6(d) Δx = 3λ/4: Δφ = 3π/2 rad.

Step-by-step solution

1. 1(a) x and t separate into factors — sin(2πx/3) times cos(120πt) — so the wave is a stationary wave.
2. 2(b) Using sin a cos b = ½[sin(a + b) + sin(a − b)]: y = 0.03 sin(120πt + 2πx/3) + 0.03 sin(120πt − 2πx/3).
3. 3Each wave has k = 2π/3 m⁻¹ and ω = 120π rad s⁻¹: λ = 2π/k = 3 m, f = ω/2π = 60 Hz, v = ω/k = 180 m s⁻¹ (opposite directions).
4. 4(c) μ = 3.0 × 10⁻²/1.5 = 0.02 kg m⁻¹, so T = μv² = 0.02 × 180² = 648 N.

Step-by-step solution

1. 1(i) In a stationary wave (the wave of Ex 14.11, referred to here): (a) every point oscillates at the source frequency 60 Hz — yes, same frequency; (b) points between two adjacent nodes are all in the same phase — yes, same phase; (c) the amplitude 0.06 |sin(2πx/3)| varies from node (0) to antinode — no, different amplitudes.
2. 2(ii) At x = 0.375 m: A(x) = 0.06 sin(2π × 0.375/3) = 0.06 sin(π/4).
3. 3A = 0.06 × 0.707 ≈ 4.2 × 10⁻² m.

Step-by-step solution

1. 1(a) Product of a pure x-function and a pure t-function — a stationary wave.
2. 2(b) y = 2√(x − vt) is a bounded function of (x − vt) — a travelling wave (pulse).
3. 3(c) Both terms carry the same argument (5x − 0.5t); the sum is a single travelling wave.
4. 4(d) Sum of two stationary waves of different wavelengths and frequencies — neither a pure travelling wave nor a pure stationary wave.

Step-by-step solution

1. 1Length L = m/μ = 3.5 × 10⁻²/4.0 × 10⁻² = 0.875 m.
2. 2(a) Fundamental on a string: f₁ = v/2L, so v = 2Lf₁ = 2 × 0.875 × 45 = 78.75 m s⁻¹.
3. 3(b) T = μv² = 4.0 × 10⁻² × (78.75)² ≈ 248 N.

Step-by-step solution

1. 1For a pipe closed at one end, successive resonances occur at lengths separated by λ/2.
2. 2λ/2 = 79.3 − 25.5 = 53.8 cm, so λ = 107.6 cm = 1.076 m.
3. 3v = fλ = 340 × 1.076 ≈ 366 m s⁻¹.

Step-by-step solution

1. 1Clamped at the middle, the centre is a node and both ends are antinodes; for the fundamental mode λ/2 = L.
2. 2λ = 2L = 2 × 1.00 = 2.00 m, and f = 2530 Hz.
3. 3v = fλ = 2530 × 2 = 5060 m s⁻¹.

Step-by-step solution

1. 1Closed pipe: fundamental f₁ = v/4L = 340/(4 × 0.20) = 425 Hz; natural frequencies are 425 Hz, 1275 Hz, 2125 Hz etc.
2. 2The 430 Hz source lies practically on the fundamental (425 Hz), so the pipe is excited in its first (fundamental) harmonic mode — nearly to resonance.
3. 3Open pipe: fundamental f₁ = v/2L = 340/0.40 = 850 Hz; harmonics are 850 Hz, 1700 Hz, 2550 Hz.
4. 4430 Hz is not an integer multiple of 850 Hz, so the same source will NOT resonate the open pipe.

Step-by-step solution

1. 1Initially |fA − fB| = 6, so fB = 330 Hz or 318 Hz.
2. 2Tension of A is reduced, which lowers fA (f ∝ √T).
3. 3If fB = 330 Hz, decreasing fA would move it further from 330, increasing the beat frequency — contradicts the observed drop to 3 Hz.
4. 4If fB = 318 Hz, decreasing fA moves it towards 318, reducing beats from 6 to 3 Hz — consistent.

Step-by-step solution

1. 1(a) At a displacement node the particles stay put, so the air is periodically compressed and rarefied — a pressure antinode; at a displacement antinode particles move freely and no pressure variation builds up — a pressure node.
2. 2(b) Bats emit high-frequency ultrasonic pulses and detect the echoes: the time delay gives distance, the Doppler shift and intensity give direction and speed, and the pattern of the reflected sound discloses the obstacle's nature and size.
3. 3(c) The two notes differ in quality (timbre): a violin note and a sitar note carry different numbers and relative intensities of overtones, even at the same fundamental frequency.
4. 4(d) Solids possess rigidity (shear modulus) in addition to bulk modulus, so they transmit both transverse and longitudinal waves; gases and liquids have no shear resistance, so only longitudinal waves can travel in them.
5. 5(e) A pulse is a sum of many frequency components; in a dispersive medium each component travels at a different speed, so the components separate and the pulse spreads or distorts.