Chapter 14: Waves Application Questions | physics Class 11 CBSE

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Chapter 14: Waves Application Questions | physics Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A steel wire of length 0.72 m has a mass of 5.0 × 10⁻³ kg and is under a tension of 60 N. Calculate the speed of transverse waves on the wire.

The linear mass density is μ = mass/length = (5.0 × 10⁻³ kg)/(0.72 m) = 6.94 × 10⁻³ kg m⁻¹.
Using the formula for transverse wave speed on a string: v = √(T/μ) = √(60 / 6.94 × 10⁻³).
v = √(8645) ≈ 93 m s⁻¹.
This speed depends only on the tension and linear mass density of the string, not on the frequency or amplitude of any wave propagating through it.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A wave travelling along a string is described by y(x, t) = 0.005 sin(80.0x − 3.0t) SI units. Determine the amplitude, wavelength, period, frequency, and wave speed.

Comparing with y = a sin(kx − ωt): amplitude a = 0.005 m = 5 mm; angular wave number k = 80.0 rad m⁻¹; angular frequency ω = 3.0 rad s⁻¹.
Wavelength λ = 2π/k = 2π/80.0 ≈ 0.0785 m = 7.85 cm. Period T = 2π/ω = 2π/3.0 ≈ 2.09 s. Frequency ν = 1/T ≈ 0.48 Hz.
Wave speed v = ω/k = 3.0/80.0 = 0.0375 m s⁻¹ (alternatively v = λν = 7.85 × 10⁻² × 0.48 ≈ 0.0375 m s⁻¹).
The wave travels in the positive x-direction because x and t appear as (kx − ωt) with a negative sign before ωt.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A bat emits ultrasonic sound of frequency 1000 kHz in air. The sound meets a water surface. Find the wavelength of (a) the reflected sound in air and (b) the transmitted sound in water. (Speed of sound: air = 340 m s⁻¹, water = 1486 m s⁻¹.)

When sound reflects from the water surface and returns into air, the frequency remains unchanged at ν = 1000 kHz = 10⁶ Hz; the wavelength in air is λair = vair/ν = 340/10⁶ = 3.4 × 10⁻⁴ m = 0.34 mm.
When sound is transmitted into water, the frequency is still ν = 10⁶ Hz (frequency is conserved across boundaries); the wavelength in water is λwater = vwater/ν = 1486/10⁶ = 1.486 × 10⁻³ m ≈ 1.49 mm.
The wavelength in water is greater than in air because the speed of sound is much higher in water (1486 m/s vs 340 m/s) while frequency is the same.
This result illustrates the general principle: when a wave crosses a boundary, frequency is conserved but wavelength changes in proportion to the speed in each medium (λ = v/ν).

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A hospital ultrasonic scanner operates at 4.2 MHz to locate tumours. The speed of sound in the tissue is 1.7 km s⁻¹. What is the wavelength of the ultrasound in the tissue? Why is such a short wavelength needed?

The wavelength of ultrasound in tissue is λ = v/ν = 1700 m s⁻¹ / (4.2 × 10⁶ Hz) = 4.05 × 10⁻⁴ m ≈ 0.4 mm.
Such a short wavelength is necessary because the resolution of any wave-based imaging system is limited by its wavelength; features smaller than λ cannot be resolved.
With λ ≈ 0.4 mm, the scanner can detect tumours and tissue structures on the millimetre scale, which is clinically significant for diagnosis.
Using audible sound (much lower frequency, longer wavelength) would give much poorer resolution, making it impossible to locate small tumours accurately within tissue.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

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

The linear mass density of the string is μ = m/L = 2.50/20.0 = 0.125 kg m⁻¹.
The speed of the transverse disturbance on the string is v = √(T/μ) = √(200/0.125) = √1600 = 40 m s⁻¹.
The time taken for the disturbance to travel the full length of the string is t = L/v = 20.0/40 = 0.5 s.
The disturbance travels as a transverse wave pulse; it is the wave (disturbance) that travels at 40 m/s, not the string itself — the string elements only oscillate about their mean positions.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 6

Applied Concept

Two sitar strings A and B playing the note 'Dha' produce 5 beats per second. The tension in string B is slightly increased, and the beat frequency drops to 3 Hz. If the frequency of A is 427 Hz, find the original frequency of B.

Since increasing tension in B raises its frequency (ν ∝ √T) and the beat frequency decreased from 5 to 3 Hz, string B is becoming closer in frequency to A — meaning B was originally at a lower frequency than A.
If νB > νA, increasing νB would increase the beat frequency |νA − νB|, not decrease it. Since beats decreased, νB < νA.
Therefore νA − νB = 5 Hz, giving νB = 427 − 5 = 422 Hz.
After increasing tension, νB increases, reducing the beat frequency to 3 Hz, which is consistent: the gap between νA = 427 Hz and the new νB = 424 Hz would be 3 Hz.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

A pipe 30.0 cm long is open at both ends. Which harmonic resonates with a 1.1 kHz source? Will the same source cause resonance if one end is closed? (Speed of sound = 330 m s⁻¹.)

For the open pipe (L = 0.30 m): νn = nv/2L = n × 330/(2 × 0.30) = 550n Hz. For n = 1: 550 Hz; n = 2: 1100 Hz. Since 1.1 kHz = 1100 Hz = ν2, the second harmonic resonates.
For the same pipe closed at one end: only odd harmonics are allowed — νn = (2n+1)v/4L = (2n+1) × 330/1.2 = (2n+1) × 275 Hz: 275 Hz, 825 Hz, 1375 Hz, 1925 Hz, ...
The frequency 1100 Hz does not appear in this sequence of odd harmonics, so no resonance is observed with the closed pipe.
This demonstrates the fundamental difference between open and closed pipes: open pipes support all harmonics, while closed pipes support only odd harmonics; the same source may resonate with one configuration but not the other.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

A transverse harmonic wave on a string is given by y(x, t) = 3.0 sin(36t + 0.018x + π/4) where x, y are in cm and t in s. Is this a travelling wave? Find its speed and direction of propagation.

Since y depends on (x, t) in the combination (0.018x + 36t), and both terms have the same sign, this is a travelling wave moving in the negative x-direction (because y = a sin(kx + ωt) moves in the −x direction).
From the equation: angular wave number k = 0.018 rad cm⁻¹ = 1.8 rad m⁻¹; angular frequency ω = 36 rad s⁻¹.
Speed of the wave: v = ω/k = 36/0.018 = 2000 cm s⁻¹ = 20 m s⁻¹.
The wave travels in the negative x-direction (i.e., from right to left) at 20 m/s, with amplitude 3.0 cm and initial phase π/4.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

A wire of length 12.0 m and mass 2.10 kg is to carry transverse waves at the speed of sound in dry air at 20°C (343 m s⁻¹). What tension must be applied to the wire?

The linear mass density of the wire is μ = m/L = 2.10/12.0 = 0.175 kg m⁻¹.
For the transverse wave speed to equal 343 m s⁻¹, use v = √(T/μ), giving T = μv² = 0.175 × (343)² = 0.175 × 117649.
T = 20588 N ≈ 2.06 × 10⁴ N.
This is an extremely large tension (roughly 2000 kg-force), illustrating why achieving sound-speed transverse waves in a solid wire requires very high tensions far beyond practical engineering limits for most materials.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

A stone dropped from the top of a 300 m tower splashes into water at the base. When is the splash heard at the top? (Speed of sound = 340 m s⁻¹, g = 9.8 m s⁻².)

Time for the stone to fall 300 m: using s = ½gt², t1 = √(2s/g) = √(2 × 300/9.8) = √61.2 ≈ 7.82 s.
Time for the sound of the splash to travel 300 m back to the top: t2 = d/v = 300/340 ≈ 0.88 s.
Total time from dropping the stone to hearing the splash at the top: t = t1 + t2 = 7.82 + 0.88 ≈ 8.7 s.
This problem illustrates that the sound wave travels back to the observer at the speed of sound (340 m/s), independent of the speed of the stone that generated it — the medium (air) alone determines the wave speed.

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Chapter sections

Chapter 14: Waves | Application Questions

A steel wire of length 0.72 m has a mass of 5.0 × 10⁻³ kg and is under a tension of 60 N. Calculate the speed of transverse waves on the wire.

A wave travelling along a string is described by y(x, t) = 0.005 sin(80.0x − 3.0t) SI units. Determine the amplitude, wavelength, period, frequency, and wave speed.

A bat emits ultrasonic sound of frequency 1000 kHz in air. The sound meets a water surface. Find the wavelength of (a) the reflected sound in air and (b) the transmitted sound in water. (Speed of sound: air = 340 m s⁻¹, water = 1486 m s⁻¹.)

A hospital ultrasonic scanner operates at 4.2 MHz to locate tumours. The speed of sound in the tissue is 1.7 km s⁻¹. What is the wavelength of the ultrasound in the tissue? Why is such a short wavelength needed?

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

Two sitar strings A and B playing the note 'Dha' produce 5 beats per second. The tension in string B is slightly increased, and the beat frequency drops to 3 Hz. If the frequency of A is 427 Hz, find the original frequency of B.

A pipe 30.0 cm long is open at both ends. Which harmonic resonates with a 1.1 kHz source? Will the same source cause resonance if one end is closed? (Speed of sound = 330 m s⁻¹.)

A transverse harmonic wave on a string is given by y(x, t) = 3.0 sin(36t + 0.018x + π/4) where x, y are in cm and t in s. Is this a travelling wave? Find its speed and direction of propagation.

A wire of length 12.0 m and mass 2.10 kg is to carry transverse waves at the speed of sound in dry air at 20°C (343 m s⁻¹). What tension must be applied to the wire?

A stone dropped from the top of a 300 m tower splashes into water at the base. When is the splash heard at the top? (Speed of sound = 340 m s⁻¹, g = 9.8 m s⁻².)

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