Waves: CBSE Class 11 Physics (NCERT)

What are Waves?

A wave is a disturbance that propagates through a medium (or even a vacuum, in the case of electromagnetic waves) without the net transfer of matter. Instead, it transfers energy and momentum. Imagine dropping a pebble into a calm pond; ripples spread outwards. The water particles themselves do not travel with the ripple; they simply oscillate up and down in their positions while the disturbance (energy) moves forward. This distinction between matter movement and energy transfer is crucial. Waves are broadly classified based on the medium they require and the direction of particle oscillation relative to wave propagation. Understanding these basic principles sets the stage for comprehending more complex wave phenomena like interference and diffraction, which you'll encounter later.

Types of Waves and Key Characteristics

Waves can be broadly categorized in several ways. Firstly, based on the medium requirement:

1. Mechanical Waves: These waves require a material medium for their propagation. They cannot travel through a vacuum. Examples include sound waves, water waves, and seismic waves. Their propagation involves the oscillation of particles of the medium, transferring energy through elastic forces.
2. Electromagnetic Waves: These waves do not require any material medium for their propagation and can travel through a vacuum. They consist of oscillating electric and magnetic fields perpendicular to each other and to the direction of wave propagation. Examples include light, radio waves, X-rays, and gamma rays.

Secondly, based on the direction of particle oscillation relative to wave propagation:

1. Transverse Waves: In these waves, the particles of the medium oscillate perpendicular to the direction of wave propagation. A classic example is a wave on a string, where the string segments move up and down, but the wave itself travels horizontally. Light waves are also transverse waves.
2. Longitudinal Waves: In these waves, the particles of the medium oscillate parallel to the direction of wave propagation. Sound waves are the best example, where compressions and rarefactions travel along the direction of sound, and air particles oscillate back and forth in the same direction.

Key Characteristics of a Wave:

• Amplitude (A): The maximum displacement of a particle from its mean position. It indicates the energy carried by the wave (Energy ∝ A²).
• Wavelength (λ): The distance between two consecutive points in the same phase (e.g., two consecutive crests or troughs in a transverse wave, or two consecutive compressions or rarefactions in a longitudinal wave).
• Frequency (f or ν): The number of oscillations or wave cycles per unit time. It is measured in Hertz (Hz).
• Time Period (T): The time taken for one complete oscillation or wave cycle. It is the reciprocal of frequency (T = 1/f).
• Wave Velocity (v): The speed at which the wave disturbance propagates through the medium. It is given by the relation: v = fλ.

The General Wave Equation and Its Applications

1. Understanding the Standard Wave Equation — A sinusoidal progressive wave propagating along the positive x-axis can be mathematically represented by the equation: y(x, t) = A sin(kx - ωt + φ). Let's break down each term: y(x, t): The displacement of a particle at position x and time t from its equilibrium position. A: Amplitude of the wave, representing the maximum displacement. k: Angular wave number (or propagation constant), defined as k = 2π/λ. It tells us how many radians of phase change occur per unit distance. ω: Angular frequency, defined as ω = 2πf = 2π/T. It tells us how many radians of phase change occur per unit time. (kx - ωt): The phase of the wave at position x and time t. φ: Initial phase angle (or phase constant), representing the phase of the wave at x = 0 and t = 0. This term is crucial when dealing with phase differences.
2. Relationship between Wave Speed, Frequency, and Wavelength — The speed of the wave, v, is related to its angular frequency ω and angular wave number k by: v = ω/k. Substituting the definitions of ω and k, we get: v = (2πf) / (2π/λ) = fλ. This fundamental relationship allows us to calculate any one of these quantities if the other two are known.
3. Worked Example 1: Extracting Wave Parameters from Equation — Consider a wave described by the equation y(x, t) = 0.05 sin(2πx - 4πt) where x and y are in meters and t is in seconds. Determine the amplitude, wavelength, frequency, and wave speed. Step 1: Compare with the general equation. The general wave equation is y(x, t) = A sin(kx - ωt + φ). Comparing this with the given equation, y(x, t) = 0.05 sin(2πx - 4πt). Step 2: Identify Amplitude (A). From comparison, A = 0.05 m. Step 3: Identify Angular Wave Number (k) and calculate Wavelength (λ). From comparison, k = 2π. We know k = 2π/λ. So, 2π = 2π/λ which implies λ = 1 m. Step 4: Identify Angular Frequency (ω) and calculate Frequency (f). From comparison, ω = 4π. We know ω = 2πf. So, 4π = 2πf which implies f = 2 Hz. Step 5: Calculate Wave Speed (v). Using v = fλ: v = (2 Hz) * (1 m) = 2 m/s. Alternatively, using v = ω/k: v = (4π) / (2π) = 2 m/s.

Exam Tips and Common Mistakes in Waves

When tackling wave problems in your CBSE Class 11 Physics exams, keep these crucial points in mind to avoid common errors:

1. Particle Velocity vs. Wave Velocity: A very frequent mistake is confusing the velocity of the wave (v = fλ) with the velocity of the particles of the medium. For a transverse wave y = A sin(kx - ωt), the wave velocity v = ω/k is constant, but the particle velocity is v_p = dy/dt = -Aω cos(kx - ωt). This particle velocity is not constant; it varies with time and position, being maximum at the mean position and zero at extreme displacements. Always be clear about which velocity the question is asking for.
2. Units: Pay close attention to units! Ensure consistency across all quantities. If wavelength is in meters, frequency in Hz, then speed will be in m/s. Radians are the standard unit for phase and angular frequency/wave number.
3. Sign Convention in Wave Equation: The sign in (kx ± ωt) determines the direction of wave propagation. (kx - ωt) represents a wave travelling in the positive x-direction, while (kx + ωt) represents a wave travelling in the negative x-direction. A common error is mixing these up, leading to incorrect direction.
4. Phase Difference: The phase difference between two particles at different positions (Δx) at the same time is Δφ = kΔx = (2π/λ)Δx. The phase difference between a single particle at different times (Δt) is Δφ = ωΔt = (2π/T)Δt. Remember 2π (or 360°) corresponds to a full cycle or wavelength. These calculations are frequently tested.
5. Understanding Stationary Waves: For stationary waves, particles at nodes have zero displacement and maximum pressure/density variation, while particles at antinodes have maximum displacement and minimum pressure/density variation. Don't confuse the characteristics of progressive and stationary waves.

Practice Questions with Solutions

• Q: 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. Find the amplitude, initial phase angle, wavelength, and frequency of the wave. Is it travelling in the positive or negative x-direction? A: Step 1: Compare the given equation y(x, t) = 3.0 sin(36t + 0.018x + π/4) with the standard form y(x, t) = A sin(ωt + kx + φ) (or A sin(kx + ωt + φ)). Step 2: Identify amplitude (A) and initial phase (φ). From comparison, Amplitude A = 3.0 cm. Initial phase angle φ = π/4 radians. Step 3: Identify angular frequency (ω) and calculate frequency (f). From comparison, ω = 36 rad/s. Since ω = 2πf, then f = ω/(2π) = 36/(2π) ≈ 5.73 Hz. Step 4: Identify angular wave number (k) and calculate wavelength (λ). From comparison, k = 0.018 rad/cm. Since k = 2π/λ, then λ = 2π/k = 2π/0.018 ≈ 349.07 cm. Step 5: Determine the direction of propagation. The equation has (ωt + kx) with positive signs for both ωt and kx. This indicates that the wave is travelling in the negative x-direction. Final answer: Amplitude = 3.0 cm, Initial phase angle = π/4 rad, Frequency = 5.73 Hz, Wavelength = 349.07 cm. The wave is travelling in the negative x-direction.
• Q: A sound wave has a frequency of 2 kHz and wavelength 35 cm. How long will it take to travel 1.5 km? A: Step 1: Convert units to SI. Frequency f = 2 kHz = 2 × 10³ Hz. Wavelength λ = 35 cm = 0.35 m. Distance d = 1.5 km = 1500 m. Step 2: Calculate the speed of the sound wave. Wave speed v = fλ = (2 × 10³ Hz) × (0.35 m) = 700 m/s. Step 3: Calculate the time taken to travel the given distance. Time t = Distance / Speed = d / v = 1500 m / 700 m/s ≈ 2.14 s. Final answer: It will take approximately 2.14 seconds to travel 1.5 km.
• Q: The equation of a wave is given by y = 10 sin(π(x - 50t)) where x and y are in cm and t is in seconds. Determine the wave velocity and maximum particle velocity. A: Step 1: Rewrite the equation in standard form. The given equation is y = 10 sin(πx - 50πt). Comparing with y = A sin(kx - ωt). Step 2: Identify A, k, and ω. Amplitude A = 10 cm. Angular wave number k = π rad/cm. Angular frequency ω = 50π rad/s. Step 3: Calculate the wave velocity (v). Wave velocity v = ω/k = (50π rad/s) / (π rad/cm) = 50 cm/s. Step 4: Calculate the maximum particle velocity (v_p_max). Particle velocity v_p = dy/dt = -Aω cos(kx - ωt). The maximum particle velocity occurs when cos(kx - ωt) = ±1, so v_p_max = Aω. v_p_max = (10 cm) × (50π rad/s) = 500π cm/s ≈ 1570.8 cm/s. Final answer: Wave velocity = 50 cm/s. Maximum particle velocity = 500π cm/s (or approximately 1570.8 cm/s).

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