NCERT Solutions for Class 11 Physics Chapter 14: Waves

Concept. A transverse disturbance travels along a stretched string with the speed v = √(T/μ), where T is the tension and μ the mass per unit length. The time to cover the length L is L/v.

Question. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If a transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Given

• Mass M = 2.50 kg; length L = 20.0 m
• Tension T = 200 N

Solution

Linear mass density:

μ = M/L = 2.50/20.0 = 0.125 kg m⁻¹

Speed of the transverse wave:

v = √(T/μ) = √(200/0.125) = √1600 = 40 m s⁻¹

Time to reach the other end:

t = L/v = 20.0/40 = 0.50 s

The disturbance takes 0.50 s to reach the other end.

Remarks

What determines the speed? Only the tension and the mass density of the string: v ∝ √T and v ∝ 1/√μ. The strength of the jerk (amplitude) and its shape do not affect the speed; a stronger jerk gives a larger displacement but the same speed. Doubling the tension would raise the speed by √2 = 1.41 and the transit time would fall to 0.35 s.

Idealisation. The string is assumed uniform, perfectly flexible, with the tension the same everywhere (the weight of the string is neglected, which is reasonable here since 200 N is much larger than the weight 2.5 × 9.8 = 24.5 N if the string is horizontal, though the sag would not be negligible in practice).

Unit check. N/(kg m⁻¹) = kg m s⁻²/(kg m⁻¹) = m² s⁻², whose square root is m s⁻¹ ✓.

Extra depth. The speed on a string is the same for a pulse and for a harmonic wave of any frequency (a non-dispersive medium). A guitar string is tuned by changing the tension: a higher tension gives a higher speed and a higher fundamental frequency ν = (1/2L)√(T/μ). The thickest strings have the largest μ and the lowest frequency.

More to know. The jerk produces a pulse that keeps its shape while travelling because the string is a non-dispersive medium. On reaching the far end the pulse is reflected (inverted if the end is fixed, upright if the end is free) and comes back to the starting end after twice the time, 1.0 s here. If the jerks are repeated at regular intervals of 1.0 s, the reflected pulses overlap with the new ones and set up a standing wave pattern, which is the basis of the stringed instruments.

Conclusion: The disturbance reaches the other end in 0.50 s.

Concept. The splash is heard after a total time made of two parts: the time t₁ for the stone to fall 300 m (free fall) and the time t₂ for the sound to travel back up 300 m at constant speed.

Question. A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top, given that the speed of sound in air is 340 m s⁻¹? (g = 9.8 m s⁻²)

Given

• h = 300 m; initial speed of the stone u = 0; g = 9.8 m s⁻²
• Speed of sound v = 340 m s⁻¹

Solution

Step 1: time of fall (from rest, h = ½gt₁²):

t₁ = √(2h/g) = √(2 × 300/9.8) = √61.2 = 7.82 s

Step 2: time for the sound to travel up the tower:

t₂ = h/v = 300/340 = 0.88 s

Step 3: total time:

t = t₁ + t₂ = 7.82 + 0.88 = 8.7 s

The splash is heard at the top 8.7 s after the stone is dropped.

Comments

• The time taken by the sound (0.88 s) is about 10 % of the total: neglecting it would give 7.8 s, an error of 11 %. This is why the time-of-fall method of measuring the depth of a well must be corrected for the travel time of sound.
• Air resistance on the stone is neglected; in practice it reduces the speed of fall, so the real fall time would be slightly longer than 7.82 s.
• The speed of sound is constant (340 m s⁻¹), independent of the height and the frequency of the splash.

Using the pond to find depths. If the time T between dropping a stone and hearing the splash in a well is measured, h satisfies T = √(2h/g) + h/v, a quadratic in √h that can be solved to give h.

Extra depth. Mixing a motion under gravity with the propagation of a wave is a standard type of problem. The two stages are independent, so their times simply add. One could also compute the speed of the stone on impact: 76.6 m s⁻¹; this is why people find a splash very loud when stones are dropped from a bridge.

Conclusion: The splash is heard 8.7 s after the stone is dropped (7.82 s + 0.88 s).

Concept. The speed of a transverse wave on a wire is v = √(T/μ), so the tension needed is T = μv², with μ the mass per unit length.

Question. A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s⁻¹?

Given

• L = 12.0 m; M = 2.10 kg
• Required v = 343 m s⁻¹

Solution

μ = M/L = 2.10/12.0 = 0.175 kg m⁻¹

v = √(T/μ) ⟹ T = μv² = 0.175 × (343)²

T = 0.175 × 1.176 × 10⁵ = 2.06 × 10⁴ N

A tension of about 2.06 × 10⁴ N (≈ 2.1 × 10⁴ N, about 2.1 tonne-weight) is needed.

Check

With T = 2.06 × 10⁴ N and μ = 0.175 kg m⁻¹, v = √(2.06 × 10⁴/0.175) = √(1.177 × 10⁵) = 343 m s⁻¹ ✓.

Is this tension possible? The tension must be borne by the wire’s cross-section. The wire’s volume is M/ρ_steel = 2.10/7800 = 2.69 × 10⁻⁴ m³, so the cross-section is A = V/L = 2.24 × 10⁻⁵ m² (diameter 5.3 mm). The stress is T/A = 2.06 × 10⁴/2.24 × 10⁻⁵ = 9.2 × 10⁸ Pa, which is well above the yield strength of ordinary mild steel (about 2.5 × 10⁸ Pa) but below the breaking stress of high-strength steel (1 – 2 × 10⁹ Pa). So the wire would have to be made of high-strength steel; the numbers are realistic only for such a wire.

Common mistake. Using the total mass (2.10 kg) rather than the mass per unit length: T = Mv² would be wrong by a factor of L.

Extra depth. In a stretched wire, the speed of transverse waves is v = √(T/μ) = √(stress/density): since μ = ρA, v = √((T/A)/ρ). For steel with stress 9.2 × 10⁸ Pa and density 7800 kg m⁻³, v = 343 m s⁻¹, as required. The maximum speed of transverse waves in a wire is limited by the breaking stress: v_max = √(σ_max/ρ) ≈ 300 – 600 m s⁻¹ for steel.

Conclusion: T ≈ 2.06 × 10⁴ N.

Concept. In v = √(γP/ρ), the speed depends on the ratio P/ρ. For an ideal gas P/ρ = RT/M (M = molar mass), so v = √(γRT/M). Any change that changes P/ρ changes the speed.

Question. Use the formula v = √(γP/ρ) to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.

(a) Independence of pressure

For a fixed mass of air at constant temperature, Boyle’s law gives P V = constant, so P/ρ = (PV)/m = constant: when the pressure is increased the density increases in the same proportion. P/ρ stays the same, so v = √(γP/ρ) is unchanged. Equivalently, for an ideal gas v = √(γRT/M) contains neither P nor ρ. (The ratio γ is also independent of pressure for a given gas.) So at a given temperature the speed of sound at the top of a mountain is the same as at sea level.

(b) Increase with temperature

At constant pressure, a rise in temperature makes the gas expand, so the density ρ decreases (ρ ∝ 1/T). Since P/ρ ∝ T, the speed v = √(γP/ρ) ∝ √T increases. Using v = √(γRT/M): v₂/v₁ = √(T₂/T₁); the speed increases by about 0.61 m s⁻¹ for each degree Celsius near 0 °C. For example, v = 331 m s⁻¹ at 0 °C and 343 m s⁻¹ at 20 °C: the ratio 343/331 = 1.036 = √(293/273) = 1.036 ✓.

(c) Increase with humidity

Humid air contains water vapour (M = 18 g mol⁻¹), which is lighter than the nitrogen (28) and oxygen (32) it replaces (the mean molar mass of dry air is 29 g mol⁻¹). At the same temperature and pressure, equal volumes of gases contain equal numbers of molecules (Avogadro’s law), so the replacement of some N₂ or O₂ molecules by H₂O molecules reduces the density of the moist air compared with dry air. Because P is unchanged, P/ρ is greater and so v = √(γP/ρ) is greater: sound travels faster in humid air than in dry air (the change of γ is small). This is why sound carries faster on a humid day and over a damp lake.

Numerical illustration for (c). At 20 °C with 100 % relative humidity, about 2.3 % of the molecules are water vapour. The mean molar mass falls from 28.96 to 28.96 − 0.023 × (28.96 − 18) = 28.71 g mol⁻¹, so v rises by about ½ × 0.9 % = 0.4 % (about 1.4 m s⁻¹).

Extra depth. Newton’s formula v = √(P/ρ) with the isothermal assumption gives 280 m s⁻¹, but Laplace’s adiabatic correction (the factor γ = 1.4) gives 331 m s⁻¹ at 0 °C. Because of the dependence on temperature, the musical instruments such as flutes and organ pipes go sharp in a warm room and flat in a cold one, since the frequency ν = v/2L (or v/4L) depends on v.

Conclusion: (a) P/ρ unchanged; (b) ρ falls as T rises so v ∝ √T; (c) moist air is less dense so v increases.

Concept. A travelling wave function must be a function of (x ± vt), but this is not enough: the displacement of a real wave must be finite (and single-valued) at all x and t. A function that blows up cannot describe the displacement of a physical medium.

Question. A travelling wave in one dimension is represented by y = f(x, t), where x and t appear in the combination x − vt or x + vt, i.e. y = f(x ± vt). Is the converse true? Examine whether the following functions of y can possibly represent a travelling wave: (a) (x − vt)²; (b) log[(x + vt)/x₀]; (c) 1/(x + vt).

The test

The converse is not true: a function of x ± vt does not always represent a wave. A wave function must also be finite for all values of x and t, because the displacement of an element cannot be infinite. Moreover, in a real wave, the displacement of elements should remain bounded as the wave passes.

Explanation

• (a) y = (x − vt)² increases without bound as x or t grows. Such a displacement is impossible for a medium; at large |x − vt| the displacement is unphysically large. A wave is a bounded disturbance.
• (b) y = log[(x + vt)/x₀] becomes infinite when x + vt → ∞ (and −∞ as x + vt → 0), so it is not finite everywhere.
• (c) y = 1/(x + vt) is infinite at x + vt = 0, i.e. at x = −vt, a point that moves with speed v. The function is finite elsewhere; but a wave must be finite at every point and at every time, so it fails the test. In the restricted domain x + vt not equal to 0 it could model a pulse, but not a wave over the whole line.

What a valid wave looks like. Examples of valid functions: a sinusoid sin(kx − ωt), a Gaussian pulse exp[−(x − vt)²/a²] or the pulse 1/[1 + (x − vt)²]; all are finite everywhere, and they approach zero (or oscillate boundedly) at large distances.

Extra depth. The shape of a pulse is determined by f; the form f(x − vt) guarantees only that the shape moves without change with speed v (a non-dispersive wave). The requirement that f be finite everywhere comes from the physical meaning of the displacement; in a wave equation language, the solution must also satisfy the wave equation ∂²y/∂t² = v²∂²y/∂x², which any twice-differentiable f(x ± vt) satisfies, so (a) and (b) are mathematically solutions of the wave equation but not physically acceptable waves.

Conclusion: None of (a), (b), (c) represents a travelling wave: they are not finite for all x and t.

Concept. When a wave crosses from one medium into another, its frequency (fixed by the source) does not change; the speed changes with the medium, and so does the wavelength λ = v/ν. A reflected wave stays in the same medium, so it has the same speed and the same λ as the incident wave.

Question. A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m s⁻¹ and in water 1486 m s⁻¹.

Given

• ν = 1000 kHz = 1.0 × 10⁶ Hz (same for incident, reflected and transmitted waves)
• v_air = 340 m s⁻¹; v_water = 1486 m s⁻¹

(a) Reflected sound

The reflected sound travels in air, so its speed is 340 m s⁻¹ and its frequency is unchanged:

λ_r = v_air/ν = 340/(1.0 × 10⁶) = 3.4 × 10⁻⁴ m = 0.34 mm

(b) Transmitted sound

The transmitted sound travels in water at 1486 m s⁻¹, with the same frequency:

λ_t = v_water/ν = 1486/(1.0 × 10⁶) = 1.486 × 10⁻³ m ≈ 1.5 mm

Key ideas. (i) Frequency is determined by the source and stays the same at a boundary. (ii) The wavelength is proportional to the speed: λ_t/λ_r = 1486/340 = 4.37, the same as the ratio of the speeds. (iii) The wavelength in water is longer than in air.

Reflection of ultrasound. Almost all the sound (about 99.9 %) is reflected at an air–water boundary because of the large mismatch of the acoustic impedances (ρv) of air (≈ 410 kg m⁻² s⁻¹) and water (≈ 1.5 × 10⁶): only a tiny fraction is transmitted. This is why a bat’s echolocation works well even for objects like water surfaces, and also why we cannot hear underwater conversations from the surface.

Extra depth. Bats use ultrasonic pulses of frequency 20 – 200 kHz (up to 1000 kHz in the problem). The wavelength must be comparable to or smaller than the size of the object to be detected: with λ = 0.34 mm a bat can detect objects as small as insects, while ordinary audible sound (λ ≈ 0.3 m at 1 kHz) would diffract around such small objects without producing a measurable echo.

Conclusion: (a) 3.4 × 10⁻⁴ m; (b) 1.49 × 10⁻³ m.

Concept. The wavelength is the speed divided by the frequency: λ = v/ν. For ultrasonic imaging the wavelength limits the smallest detail that can be resolved.

Question. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s⁻¹? The operating frequency of the scanner is 4.2 MHz.

Given

• v = 1.7 km s⁻¹ = 1.7 × 10³ m s⁻¹
• ν = 4.2 MHz = 4.2 × 10⁶ Hz

Solution

λ = v/ν = (1.7 × 10³)/(4.2 × 10⁶) = 4.05 × 10⁻⁴ m ≈ 4.0 × 10⁻⁴ m

The wavelength of the ultrasound in the tissue is about 0.40 mm (4.05 × 10⁻⁴ m).

Why this wavelength is chosen

• A wave can resolve details of a size about equal to its wavelength; a tumour or lesion of a millimetre or so can be located using λ = 0.4 mm. Audible sound (λ ≈ 0.3 m at 1 kHz) would diffract around such small objects and give no echo.
• High-frequency ultrasound can be directed as a narrow beam (the beam spreading by diffraction is about λ/D radians, tiny for a transducer of diameter D = 1 cm: 0.04 rad).
• The penetration depth decreases as the frequency is raised (absorption increases), so 2 – 10 MHz is a compromise between resolution and depth.

Echo ranging. A pulse travels to a tumour at a depth d and the echo returns after t = 2d/v. For d = 5 cm, t = 2 × 0.05/1700 = 59 μs; the scanner measures this time (and the direction of the beam) to build up a picture.

Unit check. km s⁻¹ must be converted to m s⁻¹ and MHz to Hz: 1.7 × 10³/4.2 × 10⁶ = 4.05 × 10⁻⁴ m. Forgetting the factor of 10³ gives an answer in km: 4.05 × 10⁻⁷ km, which is the same number but a wrong unit.

Extra depth. The speed in soft tissue (1540 m s⁻¹ on average, 1.7 km s⁻¹ in this question) is close to that in water, and the speed in bone is much higher (about 3500 m s⁻¹). Because the ultrasound is partly reflected at every interface where the acoustic impedance changes, a tumour appears as a boundary echo. Doppler ultrasound uses the frequency shift of waves reflected from moving blood.

Conclusion: λ = 4.05 × 10⁻⁴ m ≈ 0.40 mm.

Concept. Compare the given equation with the standard form y = a sin(ωt + kx + φ), which represents a wave travelling in the −x direction (the plus sign between ωt and kx). Then a = 3.0 cm, ω = 36 rad s⁻¹, k = 0.018 cm⁻¹ and φ = π/4.

Question. A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin(36t + 0.018x + π/4), where x and y are in cm and t in s. The positive direction of x is from left to right. (a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation? (b) What are its amplitude and frequency? (c) What is the initial phase at the origin? (d) What is the least distance between two successive crests in the wave?

(a) Travelling or stationary?

x and t appear in the combination (36t + 0.018x), i.e. y is a function of x + vt with v = 36/0.018: this is a travelling wave. (A stationary wave would have the product form f(x)g(t).) The plus sign means it travels in the negative x direction, from right to left.

v = ω/k = 36/0.018 = 2000 cm s⁻¹ = 20 m s⁻¹

(b) Amplitude and frequency

a = 3.0 cm, ν = ω/2π = 36/(2 × 3.14) = 5.73 Hz ≈ 5.7 Hz

The period is T = 1/ν = 0.175 s.

(c) Initial phase at the origin

At x = 0, t = 0 the phase is π/4 rad (45°): the initial phase angle φ = π/4. Then y(0, 0) = 3.0 sin(π/4) = 2.1 cm.

(d) Least distance between two successive crests

This is the wavelength:

λ = 2π/k = 2 × 3.14/0.018 = 349 cm = 3.49 m ≈ 3.5 m

As a check: λ = v/ν = 2000/5.73 = 349 cm ✓.

Common mistake. Taking the direction of propagation as +x because x is positive, or using ν = 36 Hz (it is ω = 36 rad s⁻¹; ν = ω/2π).

Extra depth. The direction of propagation can be seen from the phase: the phase ωt + kx = constant for a point of fixed phase, so x decreases as t increases. The crest which is now at x = 0 will be at x = −vt later, i.e. to the left. Wave speed 20 m s⁻¹ is typical of a taut string or cord.

Conclusion: (a) travelling, 20 m s⁻¹ towards −x; (b) 3.0 cm, 5.7 Hz; (c) π/4; (d) 3.5 m.

Concept. At a fixed position x the displacement of the wave is a function of time only: y = 3.0 sin(36t + 0.018x + π/4). For each position it is a sinusoid of the same amplitude and period; only the phase constant changes with x.

Question. For the wave described in Exercise 14.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in a travelling wave differ from one point to another: amplitude, frequency or phase?

Equations at the three positions

Shapes of the graphs

Each graph is a sinusoidal curve (a sine wave), with amplitude 3.0 cm and period T = 2π/36 = 0.175 s. The three curves are identical in shape and are shifted along the time axis by a very small amount: the phase difference between x = 0 and 2 cm is kΔx = 0.018 × 2 = 0.036 rad (2.1°), which corresponds to a time shift Δt = 0.036/36 = 1.0 × 10⁻³ s. The curves for x = 0, 2, 4 cm therefore almost coincide when plotted on a normal scale (see the figure, where they are drawn with their true shifts).

How does the motion differ from point to point?

• Amplitude: the same (3.0 cm) at all points: in a travelling wave, all elements oscillate with the same amplitude.
• Frequency: the same (5.7 Hz) at all points; it is the frequency of the source.
• Phase: different. The phase at x is greater by kx = 0.018x than at the origin; the element at larger x (a point on the left of the origin, since the wave travels leftwards) is ahead in phase. The phase difference is proportional to the distance: Δφ = kΔx = (2π/λ)Δx.

Therefore the oscillatory motion in a travelling wave differs from point to point only in phase, not in amplitude or frequency. (In a stationary wave, the opposite is true: all points between nodes are in the same phase but have different amplitudes.)

Extra depth. To see the phase difference clearly choose two points a quarter wavelength apart (Δx = 87 cm): then Δφ = π/2 and the graphs are shifted by T/4. Points λ apart (349 cm) move in exactly the same way, and points λ/2 apart (175 cm) move in opposite directions at every instant (Δφ = π).

Conclusion: Sinusoidal graphs, the same amplitude (3.0 cm) and period (0.175 s); they differ only in phase (0.036 rad per 2 cm).

Concept. In a harmonic wave y = a cos(ωt − kx + φ) the phase difference between two points separated by Δx is Δφ = kΔx = (2π/λ)Δx, regardless of the time.

Question. For the travelling harmonic wave y(x, t) = 2.0 cos 2π(10t − 0.0080x + 0.35), where x and y are in cm and t in s, calculate the phase difference between the oscillatory motion of two points separated by a distance of (a) 4 m, (b) 0.5 m, (c) λ/2, (d) 3λ/4.

Find k and λ

Compare with y = a cos(ωt − kx + φ). The phase is 2π(10t − 0.0080x + 0.35) = 20πt − 2π(0.0080)x + 0.7π. So

k = 2π × 0.0080 = 0.0503 cm⁻¹, λ = 2π/k = 1/0.0080 = 125 cm = 1.25 m

The phase difference for a separation Δx is Δφ = kΔx = 2π × 0.0080 × Δx (cm).

Results

• (a) 6.4π rad. Since 6.4π = 6π + 0.4π, the motion of the two points is the same as for a separation of 0.4π (72°); equivalently Δx/λ = 3.2 wavelengths.
• (b) 0.8π rad = 144°, since Δx/λ = 0.4.
• (c) π rad: the points are in opposite phase (one is at a crest while the other is at a trough).
• (d) 3π/2 rad (270°): equivalent to a lag of −π/2.

Check by ratio. Δφ = 2π(Δx/λ): (a) 2π × 3.2 = 6.4π ✓; (b) 2π × 0.4 = 0.8π ✓; (c) 2π × ½ = π ✓; (d) 2π × ¾ = 3π/2 ✓.

Units. The coefficient 0.0080 is in cm⁻¹, so Δx must be in cm. If Δx is entered in metres without conversion the answers would be 100 times smaller.

Extra depth. The phase difference between two points on a wave is independent of time. It depends on the separation in units of the wavelength. Points separated by whole wavelengths (Δφ = 2nπ) move identically; points separated by an odd number of half wavelengths (Δφ = (2n + 1)π) move in exact opposition. The 0.35 in the argument is only the initial phase and plays no part in the phase difference.

More to know. For two points on a wave separated by Δx the phase difference is the same at all times, but the time delay between their oscillations is Δt = Δx/v. With λ = 1.25 m and v = λν = 12.5 m s⁻¹ here, a separation of 0.5 m corresponds to a delay of 0.04 s, i.e. 40 % of the period 0.1 s: this is the same 0.4 of a cycle (0.8π) found above. Thus phase difference and time delay are two descriptions of the same fact, Δφ = ωΔt = 2πΔx/λ.

Conclusion: (a) 6.4π rad; (b) 0.8π rad; (c) π rad; (d) 3π/2 rad.

Concept. A displacement of the form y = A sin(kx) cos(ωt), in which x and t occur as separate factors, is a stationary wave. By the trigonometric identity sin A cos B = ½[sin(A + B) + sin(A − B)] it is the superposition of two identical waves of half the amplitude travelling in opposite directions.

Question. The transverse displacement of a string (clamped at both ends) is y(x, t) = 0.06 sin(2πx/3) cos(120πt), where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10⁻² kg. (a) Does the function represent a travelling wave or a stationary wave? (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What are the wavelength, frequency and speed of each wave? (c) Determine the tension in the string.

(a) Type of wave

The displacement is a product of a function of x only, sin(2πx/3), and a function of t only, cos(120πt): it represents a stationary (standing) wave. The nodes are at sin(2πx/3) = 0, i.e. x = 0, 1.5 m, 3.0 m, …; the clamped ends x = 0 and x = 1.5 m are nodes, as required.

(b) Component waves

Using sin A cos B = ½[sin(A + B) + sin(A − B)] with A = 2πx/3 and B = 120πt:

y = 0.03 sin(2πx/3 + 120πt) + 0.03 sin(2πx/3 − 120πt)

The first term is a wave travelling in the −x direction and the second in the +x direction, each of amplitude 0.03 m. Comparing with y = a sin(kx ± ωt): k = 2π/3 m⁻¹ and ω = 120π rad s⁻¹, so for each wave

λ = 2π/k = 2π/(2π/3) = 3 m, ν = ω/2π = 120π/2π = 60 Hz, v = νλ = 60 × 3 = 180 m s⁻¹

(c) Tension

μ = M/L = (3.0 × 10⁻²)/1.5 = 2.0 × 10⁻² kg m⁻¹

v = √(T/μ) ⟹ T = μv² = 2.0 × 10⁻² × (180)² = 648 N

Which mode? The string has length L = 1.5 m = λ/2, so this is the fundamental mode (n = 1): ν₁ = v/2L = 180/3.0 = 60 Hz ✓. The antinode is at the centre x = 0.75 m, where the amplitude is 0.06 m.

Extra depth. The standing wave of the string is the superposition of waves reflected at the two ends: the wave travelling to the right and its reflection from the clamped end (with a phase change of π) interfere and produce nodes at both ends. The stationary wave carries no net energy; the energy is shuttled between kinetic and potential forms, with nodes at fixed locations.

Conclusion: (a) stationary; (b) amplitude 0.03 m, λ = 3 m, ν = 60 Hz, v = 180 m s⁻¹; (c) T = 648 N.

Concept. In a stationary wave, every element of the medium oscillates with the same frequency; the amplitude depends on the position (maximum at antinodes, zero at nodes), and all elements between two successive nodes are in phase.

Question. (i) For the wave on a string described in Exercise 14.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?

(i) Frequency, phase, amplitude

The string is described by y = 0.06 sin(2πx/3) cos(120πt): an element at position x has displacement y = A(x) cos(120πt) with A(x) = 0.06 sin(2πx/3).

• (a) Frequency: yes, the same for all points. Every element executes SHM with the time factor cos(120πt): ν = 60 Hz. (Only the point at a node has no oscillation at all.)
• (b) Phase: not the same. Within a loop (between two successive nodes) all points are in the same phase because A(x) has the same sign, but the points in neighbouring loops are in opposite phase (A(x) changes sign at every node). Here the string (0 to 1.5 m) has a single loop, so all points of the string between the clamped ends oscillate in phase (they reach the maximum displacements at the same instants); the phase would be reversed in the next loop beyond x = 1.5 m.
• (c) Amplitude: not the same. It is |0.06 sin(2πx/3)|: zero at the ends (nodes) and maximum, 0.06 m, at the centre x = 0.75 m.

(ii) Amplitude at x = 0.375 m

A = 0.06 sin(2π × 0.375/3) = 0.06 sin(π/4) = 0.06 × 0.707 = 0.0424 m

The amplitude of the point 0.375 m from an end is about 0.042 m = 4.2 cm. (This point is half-way between a node and the antinode, i.e. λ/8 from the node.)

Contrast with a travelling wave. In a travelling wave all particles have the same amplitude but different phases; in a stationary wave, all particles between two nodes have the same phase but different amplitudes.

Extra depth. The amplitude function sin(2πx/3) vanishes at x = 0 and x = 1.5 m (the clamped ends) and is largest at the centre; the string vibrates as a single loop (the fundamental). In the second harmonic (3 times as many nodes), there would be a node at the centre and two loops in opposite phase.

Conclusion: (i) (a) yes, (b) no for different loops (same within the loop), (c) no; (ii) 0.042 m.

Concept. A travelling wave has the form f(x ± vt) and is finite at all x and t. A stationary wave is a product of a function of x and a function of t. A function that is neither is “none at all”.

Question. Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all: (a) y = 2 cos(3x) sin(10t); (b) y = 2√(x − vt); (c) y = 3 sin(5x − 0.5t) + 4 cos(5x − 0.5t); (d) y = cos x sin t + cos 2x sin 2t.

Analysis

Details

(c) Write 3 sin θ + 4 cos θ = 5 sin(θ + α), with θ = 5x − 0.5t, where 5 cos α = 3, 5 sin α = 4, α = 53.1° = 0.927 rad. Then y = 5 sin(5x − 0.5t + 0.927): a harmonic travelling wave towards +x with amplitude 5, k = 5, ω = 0.5 and speed v = ω/k = 0.1 (in the units of x and t), as sin A and cos A of the same argument add to give a single sinusoid.

(d) Each term is separable and is a stationary wave (a normal mode with nodes at fixed positions). Their sum is also periodic and stationary in the sense that no wave pattern propagates; the sum is a combination of the modes with ω = 1 and ω = 2 (the second a harmonic of the first). It is not a single harmonic standing wave, because the nodes of the two terms do not coincide, but it is a standing-wave type of motion: no travelling pattern.

(b) Although the argument is of the form x − vt, the function √(x − vt) is imaginary for x < vt and unbounded for large x − vt: it cannot represent a physical displacement.

Extra depth. In (c) and similar problems, combine terms of the same argument using R sin(θ + α). The result of two harmonic waves of the same frequency and wavenumber travelling in the same direction is again a travelling wave of the same kind (Section 14.5, phase difference π/2: amplitude √(3² + 4²) = 5).

Conclusion: (a) stationary; (b) none at all; (c) travelling; (d) stationary (superposition of two stationary-wave modes).

Concept. For a wire fixed at both ends the fundamental has λ = 2L, so ν₁ = v/2L. The length follows from the mass and the linear density: L = M/μ. The tension then follows from T = μv².

Question. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10⁻² kg and its linear mass density is 4.0 × 10⁻² kg m⁻¹. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

Given

• ν₁ = 45 Hz; M = 3.5 × 10⁻² kg; μ = 4.0 × 10⁻² kg m⁻¹

Solution

Length of the wire:

L = M/μ = (3.5 × 10⁻²)/(4.0 × 10⁻²) = 0.875 m

(a) Speed of the transverse wave. In the fundamental mode L = λ/2, so λ = 2L = 1.75 m:

v = νλ = 45 × 1.75 = 78.75 m s⁻¹ ≈ 79 m s⁻¹

(b) Tension:

T = μv² = 4.0 × 10⁻² × (78.75)² = 4.0 × 10⁻² × 6201.6 = 248 N ≈ 2.5 × 10² N

Check. ν₁ = (1/2L)√(T/μ) = (1/1.75)√(248/0.04) = (1/1.75)(78.74) = 45.0 Hz ✓.

Overtones. The higher harmonics are 90 Hz (second), 135 Hz (third), … so all integral multiples of 45 Hz.

Extra depth. The fundamental frequency of a string depends on length, tension and mass per unit length: ν₁ = (1/2L)√(T/μ). Doubling the tension raises the pitch by a factor √2; halving the length (pressing the string against a fret) doubles the frequency (an octave). The frequency does not depend on how hard the string is plucked.

More to know. The tension of 248 N corresponds to a weight of about 25 kg. In a real instrument such as a sitar or a violin the string’s tension is of the order of 50 – 100 N, while its linear density is a few 10⁻³ kg m⁻¹, so the fundamental of 45 Hz here is a low bass note; a thin high string has a much smaller μ and therefore a much higher frequency at a similar tension. The overtones of a plucked string are the integral multiples of the fundamental: this is why the sounds of strings are harmonic and musical.

Conclusion: (a) v ≈ 79 m s⁻¹; (b) T ≈ 2.5 × 10² N.

Concept. A tube closed at one end by the piston and open at the other resonates when its length L is an odd multiple of λ/4: L = λ/4, 3λ/4, 5λ/4 … Two successive resonance lengths therefore differ by λ/2, whatever the end correction (which is the same for both and cancels).

Question. A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed-frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

Given

• Frequency of the fork ν = 340 Hz
• Resonance lengths L₁ = 25.5 cm and L₂ = 79.3 cm
• Air column closed at the piston end and open at the other

Solution

The consecutive resonances of a closed-end pipe occur at L₁ = λ/4 and L₂ = 3λ/4 (if 25.5 cm is the first resonance), so

L₂ − L₁ = λ/2 ⟹ λ = 2(79.3 − 25.5) = 2 × 53.8 = 107.6 cm = 1.076 m

v = νλ = 340 × 1.076 = 365.8 m s⁻¹ ≈ 3.66 × 10² m s⁻¹

The speed of sound at the temperature of the experiment is about 366 m s⁻¹.

Why the difference of lengths is used

If we used L₁ = λ/4 alone, v = 4νL₁ = 4 × 340 × 0.255 = 347 m s⁻¹, which differs from 366 m s⁻¹: the first resonance is displaced by the end correction (the antinode lies slightly beyond the open end, at a distance ≈ 0.6 r from the tube’s mouth). The difference L₂ − L₁ eliminates this correction, and is therefore the preferred method. With the end correction e: L₁ + e = λ/4 and L₂ + e = 3λ/4, giving λ/2 = L₂ − L₁ (independent of e) and e = λ/4 − L₁ = 26.9 − 25.5 = 1.4 cm.

Temperature. v = 366 m s⁻¹ is higher than 343 m s⁻¹ at 20 °C: using v = 331 + 0.61t gives t ≈ 57 °C; the temperature of the experiment (or the end correction neglected in the problem) is therefore not room temperature. The given numbers are as in the textbook.

Extra depth. The resonance-tube method is the standard laboratory way to measure the speed of sound: a tuning fork of known frequency is held above a tube of water whose level is lowered; the loudness peaks when the air column is λ/4, 3λ/4, … The position of the antinode (end correction e ≈ 0.3 d for a tube of diameter d) is corrected for by taking the difference of two resonance lengths.

Conclusion: λ = 107.6 cm, v = νλ ≈ 3.66 × 10² m s⁻¹.

Concept. A rod clamped at its middle has a displacement node at the middle (the clamp) and displacement antinodes at the free ends. In the fundamental mode the distance from node to antinode is λ/4 on each side, so the length is L = λ/2.

Question. A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel?

Given

• Length L = 100 cm = 1.00 m, clamped at its midpoint
• Fundamental frequency ν = 2.53 kHz = 2530 Hz

Solution

In the fundamental mode, the clamped middle is a node (N) and the two free ends are antinodes (A), so the pattern is A–N–A: each half of the rod, of length L/2, is a quarter wavelength:

L/2 = λ/4 ⟹ λ = 2L = 2.00 m

v = νλ = 2530 × 2.00 = 5060 m s⁻¹ ≈ 5.06 km s⁻¹

The speed of sound in steel is about 5.06 km s⁻¹.

Check against a standard value

From v = √(Y/ρ) with Y = 2.0 × 10¹¹ Pa and ρ = 7800 kg m⁻³: v = √(2.56 × 10⁷) = 5060 m s⁻¹ ✓. The standard value (5960 m s⁻¹) quoted for bulk steel is higher, since the rod value uses Young’s modulus, not the bulk expression.

Overtones. The next mode that has a node at the middle is the third harmonic (λ = 2L/3), because the middle must remain a node: only odd harmonics 2.53, 7.59, 12.65 kHz, … are possible in a rod clamped at its centre.

Common mistake. Taking L = λ/4 or L = λ, i.e. using the wrong boundary condition. A rod clamped at the middle has two free ends, which are antinodes; the middle is a node.

Extra depth. The longitudinal vibration of a rod is used in ultrasonic generators (magnetostrictive or piezoelectric rods): the rod is clamped at the middle (a node, so the clamp does not damp the vibration) and its length is chosen to be half a wavelength of the desired frequency. Kundt’s tube measures the speed of sound in a rod by the dust pattern that the rod’s vibration forms in a glass tube.

More to know. In a longitudinal wave in a rod the speed is v = √(Y/ρ) and depends only on the material. With the same rod clamped at one end instead of the middle (A at the free end, N at the clamp) the fundamental would be v/4L = 1.27 kHz, half of the 2.53 kHz found here for the middle-clamped rod. Measuring the fundamental frequency of a rod is a simple way to obtain Young’s modulus: Y = ρv² = 7800 × (5060)² = 2.0 × 10¹¹ Pa for the steel here.

Conclusion: v = 5.06 km s⁻¹ (5060 m s⁻¹).

Concept. A pipe closed at one end has a node at the closed end and an antinode at the open end, so L = (2n − 1)λ/4 and ν_n = (2n − 1)v/4L: only odd harmonics. A pipe open at both ends has ν_n = nv/2L: all harmonics.

Question. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s⁻¹)

Given

• L = 20 cm = 0.20 m; v = 340 m s⁻¹; source frequency 430 Hz

Closed pipe

The frequencies of the closed pipe are ν = (2n − 1)v/4L = (2n − 1) × 340/(4 × 0.20) = (2n − 1) × 425 Hz:

The 430 Hz source is (nearly) equal to the fundamental frequency 425 Hz of the closed pipe, so it excites the first harmonic (fundamental mode). (Strictly 430 Hz is slightly higher, within 1.2 %, and an exact resonance would need a source of 425 Hz; given the approximation of the end effect and of the speed of sound, 430 Hz resonates with the fundamental.)

Open pipe (both ends open)

The frequencies are ν = nv/2L = n × 340/(2 × 0.20) = n × 850 Hz: 850, 1700, 2550 Hz, … The source frequency 430 Hz is not equal to any of these (it is about half of the fundamental of the open pipe), so the pipe is not in resonance.

Rule. The fundamental of a closed pipe is half that of an open pipe of the same length, v/4L versus v/2L, so the closed pipe sounds an octave lower.

Extra depth. An organ with closed pipes needs only half the length for the same pitch: this is why a flute (open tube) and a clarinet (closed at the mouthpiece end) have different tubes. The resonance with a source of 430 Hz is sharp when the pipe is excited at its natural frequency; for slightly different frequencies the amplitude is smaller. A real pipe has an end correction that lowers the resonant frequency slightly, which brings 425 Hz closer to a source of 430 Hz in actual experiments.

More to know. The “end correction” of a real open pipe makes its effective length slightly longer than the geometric length, by about 0.6 times the pipe radius at each open end. For a closed pipe only the open end has this correction. For a narrow pipe of 20 cm with a radius of 1 cm this lowers the resonant frequency from 425 Hz to about 412 Hz in practice. In the idealised problem the end effect is ignored, so the answer is the fundamental mode.

Conclusion: The closed pipe resonates in its fundamental (first harmonic); the open pipe (850, 1700, … Hz) does not.

Concept. The beat frequency is the difference of the two frequencies, so ν_B = ν_A ± 6 Hz. Lowering the tension of A lowers its frequency (ν ∝ √T); the change of the beat frequency then tells which sign is correct.

Question. Two sitar strings A and B playing the note “Ga” are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

Solution

Beat frequency 6 Hz means ν_B = 324 ± 6 = 330 Hz or 318 Hz.

Reducing the tension of A reduces its frequency (ν ∝ √T), so ν_A becomes somewhat less than 324 Hz.

• If ν_B were 330 Hz: lowering ν_A below 324 Hz would increase the difference (330 − ν_A) to more than 6 Hz, so the beat frequency would increase. This contradicts the observation.
• If ν_B is 318 Hz: lowering ν_A from 324 Hz towards 318 Hz decreases the difference (ν_A − 318) below 6 Hz: for example, ν_A = 321 Hz gives 3 beats per second ✓. The observation fits.

Therefore ν_B = 318 Hz.

After the adjustment. The new frequency of A is 321 Hz (3 Hz below its original 324 Hz), so that the tension was reduced by (321/324)² − 1 = −1.8 %. Further lowering ν_A to 318 Hz would remove the beats altogether (the strings are then in tune).

Extra depth. This is the method used by musicians to tune strings: tune a string against a reference note and adjust its tension until the beats slow down and finally disappear. The decision rule is: if loosening a string (lowering its pitch) reduces the beat frequency, the string was originally higher than the reference; if the beats increase, it was lower.

More to know. The 6 Hz beats are heard as a “wow–wow” modulation of loudness six times every second. A change of tension of a few per cent changes the pitch by about half as much: Δν/ν = ½ ΔT/T. Reducing the frequency of A from 324 Hz to 321 Hz is a change of −0.93 %, which needs ΔT/T = −1.9 %: a very small adjustment of the peg. Sitar and violin players tune by exactly this kind of slow beating, listening until the beats disappear.

More to know. The result also works if A’s frequency were lowered by 6 Hz or more: if ν_A fell to 318 Hz the beats would vanish and if it fell further they would reappear and then increase. In the problem the tension change is “slight”, so the decrease of ν_A is less than 6 Hz and the reasoning above (a decrease in the beat frequency from 6 Hz to 3 Hz) is unambiguous. If the beat frequency had instead increased (to 9 Hz), the answer would be ν_B = 330 Hz.

Conclusion: ν_B = 318 Hz.

Concept. These are conceptual applications of: the relation between displacement and pressure in a longitudinal wave, ultrasound reflection and echolocation, the timbre of a note, the elastic moduli needed for transverse and longitudinal waves, and dispersion.

Question. Explain why (or how): (a) in a sound wave, a displacement node is a pressure antinode and vice versa; (b) bats can ascertain distances, directions, nature and sizes of obstacles without any “eyes”; (c) a violin note and a sitar note may have the same frequency, yet we can distinguish between the two notes; (d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases; (e) the shape of a pulse gets distorted during propagation in a dispersive medium.

(a) Displacement node ⟺ pressure antinode

At a displacement node the particles of the medium are not displaced, but the layers on either side move towards or away from it in opposite directions (the particles on the two sides have displacements of opposite sign and maximum magnitude), so the node is a point of maximum compression and rarefaction: the pressure (and density) variation is greatest. At a displacement antinode the particles on both sides move in the same direction by the same amount, so there is no relative compression: the pressure variation is zero. Mathematically if s = a sin(kx) cos ωt, the pressure change is ΔP = −B ∂s/∂x = −Bka cos(kx) cos ωt: it is maximum where the displacement is zero and vice versa (a quarter wavelength shift). The closed end of a pipe is thus a displacement node and a pressure antinode; the open end is a displacement antinode and a pressure node.

(b) Bats and echolocation

A bat emits ultrasonic pulses (20 – 100 kHz or more) and listens to the echoes. The time delay of the echo gives the distance (d = vt/2); the difference between the arrival times and intensities at the two ears gives the direction; the strength and frequency content (including the Doppler shift) of the echo give the nature and size of the object (an insect, a wall or a leaf) and its motion; the short wavelength (a few mm) allows it to detect objects about as small as the wavelength. Thus the bat can navigate and find prey in darkness.

(c) Violin and sitar notes of the same frequency

A musical note has three attributes: pitch (frequency), loudness (amplitude) and timbre or quality (waveform). A violin and a sitar produce the same fundamental frequency but different combinations of overtones (harmonics) of different relative amplitudes, hence different waveforms. The ear detects these differences of harmonic content as the difference of timbre, and we distinguish the two notes.

(d) Transverse and longitudinal waves in solids and gases

A transverse wave involves shear deformation (layers sliding over each other), which requires a restoring force from a shear modulus. Solids have a shear modulus and so support both transverse and longitudinal waves. Gases (and liquids) have no shear modulus (a fluid cannot resist shear), so they cannot support transverse waves, but they have a bulk modulus (resistance to compression), so longitudinal waves (sound) can propagate in them.

(e) Pulse distortion in a dispersive medium

A pulse of any shape is a superposition of harmonic waves of many different frequencies (Fourier components). In a dispersive medium the speed of a harmonic wave depends on its frequency (or wavelength), so the components travel with different speeds and their relative phases change as the pulse advances: the shape of the pulse is distorted (usually broadened). In a non-dispersive medium all components travel with the same speed and the shape is preserved. Example: light of different colours travels with different speeds in glass (a prism); and waves on water.

Extra depth. The speed of sound in air does not depend on frequency (to a good approximation), so a pulse of sound in the open air keeps its shape; this is why speech is intelligible over long distances. In contrast, on the surface of water the speed depends on the wavelength, and waves in optical fibres are dispersed, which limits the data rate: engineers use special fibres to reduce dispersion.

Conclusion: All five parts are explained above: node ⟺ pressure antinode; echolocation; timbre; shear modulus; dispersion.