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Chapter 12: Waves (Numerical Problems)

Download free solved numericals covering wavelength calculations λ = L/n (total length divided by number of waves), wave equation v = fλ, relationship f = 1/T (frequency and time period inversely related), electromagnetic wave speed in vacuum c = 3.0 × 10⁸ m/s, wavelength of red light (700 nm = 700 × 10⁻⁹ m) gives frequency f = c/λ ≈ 4.29 × 10¹⁴ Hz, FM radio signals (90 MHz = 90 × 10⁶ Hz) gives wavelength λ = c/f = 3.33 m, ripple tank experiments (wave speed = distance/time, wavelength from crest to trough distance = λ/2), stretched string problems (10 waves on 20 m string → λ = 2 m, frequency f = v/λ), unit conversions (nanometers to meters, Megahertz to Hz), and crest to trough distance equals half wavelength - strictly according to FBISE 2026 SLOs.

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Chapter Overview & SLOs

How is the wavelength determined from the physical length of a medium? The wavelength (λ) is calculated by dividing the total length (L) of the string or tank by the number of waves (n) produced: λ = L/n.

What is the fundamental relationship between wave speed, frequency, and wavelength? The speed of a wave (v) is the product of its frequency (f) and wavelength (λ), expressed by the universal wave equation v = fλ.

How are frequency and time period related? Frequency and time period are inversely related: T = 1/f. As frequency increases, time period decreases.

What is the speed of electromagnetic waves in a vacuum? For electromagnetic waves (light, radio waves) in a vacuum, the speed is constant: c ≈ 3.0 × 10⁸ m/s.

Example 1 - Stretched string: If 10 waves are produced on a 20 m string, the wavelength is 20/10 = 2 m. If the wave speed is 12 m/s, the frequency is f = v/λ = 12/2 = 6 Hz.

Example 2 - Ripple tank: For a 1.5 m tank where waves take 2 s to cross, the speed is v = distance/time = 1.5/2 = 0.75 m/s. If the distance between a crest and trough is 10 cm, the wavelength is 20 cm (0.2 m) because crest to trough = λ/2. Resulting frequency f = v/λ = 0.75/0.2 = 3.75 Hz, and time period T = 1/f = 1/3.75 = 0.267 s.

Example 3 - Red light frequency: For red light with a wavelength of 700 nm (700 × 10⁻⁹ m) in a vacuum, the frequency is f = c/λ = (3.0 × 10⁸)/(700 × 10⁻⁹) ≈ 4.29 × 10¹⁴ Hz.

Example 4 - FM radio wavelength: For an FM station transmitting at 90 MHz (90 × 10⁶ Hz), the wavelength is λ = c/f = (3.0 × 10⁸)/(90 × 10⁶) = 3.33 m.

Important notes:

  • The distance between a successive crest and trough is half a wavelength (λ/2).
  • Unit conversions: 1 nm = 10⁻⁹ m, 1 MHz = 10⁶ Hz.

These notes are strictly aligned with the Student Learning Outcomes (SLOs) for the FBISE 2026 annual examination.

  • How do we calculate wavelength and frequency on a stretched string? If 10 waves are produced on a 20 m string, the wavelength is 20/10 = 2 m; if the speed is 12 m/s, the frequency is $f = 12/2 = 6\text{ Hz}$.
  • How are ripple tank dimensions used to find wave properties? For a 1.5 m tank where waves take 2 s to cross, the speed is 0.75 m/s; if the distance between a crest and trough is 10 cm, the wavelength is 20 cm (0.2 m), resulting in a frequency of 3.75 Hz and a time period of 0.267 s.
  • How do we find the frequency of light from its wavelength? For red light with a wavelength of 700 nm ($700 \times 10^{-9}\text{ m}$) in a vacuum, the frequency is calculated using $f = c/\lambda$, resulting in approximately $4.29 \times 10^{14}\text{ Hz}$.
  • How is the wavelength of radio signals determined? For an FM station transmitting at 90 MHz ($90 \times 10^6\text{ Hz}$), the wavelength is found by dividing the speed of light by the frequency ($\lambda = c/f$), which equals 3.33 m.

Frequently Asked Questions (FAQ)

1. Are these Class 10 Physics notes based on the latest FBISE syllabus for 2026?
Yes, these notes are strictly designed according to the Student Learning Outcomes (SLO) provided by the Federal Board (FBISE) for the 2026 academic year. We regularly update our content to match the latest curriculum changes and exam patterns.

2. Do these Physics 12 notes include solved exercise questions and diagrams?
Absolutely. These notes contain comprehensive solutions to all textbook exercise questions, including Multiple Choice Questions (MCQs), Short Questions, and detailed Long Questions. We also include labeled diagrams and key definitions to help you secure maximum marks in your board exams.

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