🎵 Chapter 13: Sound – Numerical Problems
Prepared by Muhammad Tayyab, Subject Specialist Physics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).
📖 What's Inside: This section covers numerical problems from Chapter 13 Sound including frequency calculation using wave equation, time period, echo depth using SONAR formula, and distance calculation. Each problem is presented with Given Data, To Find, and step-by-step Solution as per the official PECTAA 2026 Physics curriculum. Perfect for Punjab Boards (Lahore, Gujranwala, Multan, etc.) and all BISE boards across Pakistan.
📚 Related Resources – Chapter 13: Sound
Sound covers waves, echo, ultrasound, SONAR, pitch, loudness, and acoustics. Includes solved examples and numerical problems.
📐 Numerical Problems & Solutions (PECTAA 2026)
13.1 Frequency from Wave Speed and Wavelength (Audibility Check)
At a particular temperature, the speed of sound in air is 333 ms⁻¹. If the wavelength of a sound wave is 3 cm, calculate the frequency of the sound wave. Is this frequency in the audible range of the human ear?
Given Data
\(Wave\ speed\)\(= v = 333\ \text{ms}^{-1}\)
\(Wavelength\)\(= \lambda = 3\ \text{cm} = 3 \times 10^{-2}\ \text{m} = 0.03\ \text{m}\)
To Find
\(Frequency = f = ?\)
\(Audibility\ of\ wave = ?\)
Solution
By using the wave equation:
\[\begin{aligned} v &= f\lambda \\ 333 &= f \times 0.03 \\ \frac{333}{0.03} &= f \\ 11000 &= f \\ f &= 11000\ \text{Hz} \\ f &= 11 \times 10^{3}\ \text{Hz} \\ \boldsymbol{f} &= \boldsymbol{11\ \text{kHz}} \end{aligned}\]
Since the audible frequency range for human ears is \(20\ \text{Hz}\) to \(20\ \text{kHz}\), the calculated frequency of \(11\ \text{kHz}\) lies within this range and is audible.
✅ \(f = 11\ \text{kHz}\) — Audible
13.2 Frequency and Time Period from Chimes
A clock chimes 48 times in 1 minute. Calculate the frequency and period of the chimes.
Given Data
\(Number\ of\ chimes\)\(= n = 48\)
\(Time\)\(= t = 1\ \text{minute} = 60\ \text{s}\)
To Find
\(Frequency = f = ?\)
\(Time\ period = T = ?\)
Solution
By using the formula for frequency:
\[\begin{aligned} f &= \frac{n}{t} \\ f &= \frac{48}{60} \\ f &= 0.8\ \text{Hz} \end{aligned}\]
Now, by using the formula for time period:
\[\begin{aligned} T &= \frac{1}{f} \\ T &= \frac{1}{0.8} \\ \boldsymbol{T} &= \boldsymbol{1.25\ \text{s}} \end{aligned}\]
✅ \(f = 0.8\ \text{Hz}\) | \(T = 1.25\ \text{s}\)
13.3 Frequency of Car Horn
A car horn emits a sound with a wavelength of 0.5 m. If the speed of sound in air is 340 ms⁻¹, calculate the frequency of the sound produced by the car's horn.
Given Data
\(Wavelength\)\(= \lambda = 0.5\ \text{m}\)
\(Speed\ of\ sound\)\(= v = 340\ \text{ms}^{-1}\)
To Find
\(Frequency = f = ?\)
Solution
By using the wave equation:
\[\begin{aligned} v &= f\lambda \\ 340 &= f \times 0.5 \\ \frac{340}{0.5} &= f \\ 680 &= f \\ \boldsymbol{f} &= \boldsymbol{680\ \text{Hz}} \end{aligned}\]
✅ \(f = 680\ \text{Hz}\)
13.4 Time Period from Wave Speed and Wavelength
A wave travels with a speed of 500 ms⁻¹ and has a wavelength of 2 metres. Calculate the time period of the wave.
Given Data
\(Wave\ speed\)\(= v = 500\ \text{ms}^{-1}\)
\(Wavelength\)\(= \lambda = 2\ \text{m}\)
To Find
\(Time\ period = T = ?\)
Solution
By using the wave equation:
\[\begin{aligned} v &= f\lambda \\ 500 &= f \times 2 \\ \frac{500}{2} &= f \\ 250 &= f \\ f &= 250\ \text{Hz} \end{aligned}\]
Now, by using the formula for time period:
\[\begin{aligned} T &= \frac{1}{f} \\ T &= \frac{1}{250} \\ T &= 4 \times 10^{-3}\ \text{s} \\ \boldsymbol{T} &= \boldsymbol{0.004\ \text{s}} \end{aligned}\]
✅ \(T = 0.004\ \text{s}\)
13.5 SONAR – Depth of Sea from Echo Time
A research boat sends a sound wave straight to the seabed and receives the echo 2 seconds later. The speed of sound in seawater is 1600 ms⁻¹. Find the depth of the sea at this position.
Given Data
\(Total\ time\ for\ echo\)\(= t = 2\ \text{s}\)
\(Speed\ of\ sound\ in\ water\)\(= v = 1600\ \text{ms}^{-1}\)
To Find
\(Depth\ of\ the\ sea = d = ?\)
Solution
By using the SONAR distance formula:
\[\begin{aligned} Distance &= \frac{Speed\ of\ Sound \times Time}{2} \\[6pt] d &= \frac{1600 \times 2}{2} \\[4pt] \boldsymbol{d} &= \boldsymbol{1600\ \text{m}} \end{aligned}\]
✅ \(d = 1600\ \text{m}\)
13.6 Echo – Distance of Mountain
A person claps his hands near a mountain and hears the echo after 6 seconds. If the speed of sound is 343 ms⁻¹, what is the distance of the mountain from the person?
Given Data
\(Total\ time\ for\ echo\)\(= t = 6\ \text{s}\)
\(Speed\ of\ sound\ in\ air\)\(= v = 343\ \text{ms}^{-1}\)
To Find
\(Distance\ of\ the\ mountain = d = ?\)
Solution
By using the single-way distance formula:
\[\begin{aligned} Distance &= \frac{Speed\ of\ Sound \times Time}{2} \\[6pt] d &= \frac{343 \times 6}{2} \\[6pt] d &= \frac{2058}{2} \\[4pt] \boldsymbol{d} &= \boldsymbol{1029\ \text{m}} \end{aligned}\]
✅ \(d = 1029\ \text{m}\)
📘 Examples with Solutions
Ex 13.1 Frequency from Speed and Wavelength
Calculate the frequency of a sound wave of speed 340 ms⁻¹ and wavelength 2.0 m.
Given Data
\(Speed\ of\ sound\)\(= v = 340\ \text{ms}^{-1}\)
\(Wavelength\)\(= \lambda = 2.0\ \text{m}\)
To Find
\(Frequency = f = ?\)
Solution
By using the wave equation:
\[\begin{aligned} v &= f\lambda \\ 340 &= f \times 2.0 \\ \frac{340}{2.0} &= f \\ 170 &= f \\ \boldsymbol{f} &= \boldsymbol{170\ \text{Hz}} \end{aligned}\]
✅ \(f = 170\ \text{Hz}\)
Ex 13.2 Distance of Lightning Cloud
A flash of lightning is seen 2 seconds before the thunder is heard. How far away is the cloud in which the flash has occurred? (Speed of sound = 332 ms⁻¹)
Given Data
\(Time\)\(= t = 2\ \text{s}\)
\(Speed\ of\ sound\)\(= v = 332\ \text{ms}^{-1}\)
To Find
\(Distance\ of\ the\ cloud = S = ?\)
Solution
By using the distance formula:
\[\begin{aligned} S &= vt \\ S &= (332)(2) \\ \boldsymbol{S} &= \boldsymbol{664\ \text{m}} \end{aligned}\]
✅ \(S = 664\ \text{m}\)
Ex 13.3 SONAR – Distance to Seafloor
A sonar device on a ship sends out a sound pulse that returns after 3.2 seconds. If the speed of sound in water is 1600 ms⁻¹, what is the distance from the ship to the seafloor?
Given Data
\(Time\ for\ sound\ pulse\ to\ return\)\(= t = 3.2\ \text{s}\)
\(Speed\ of\ sound\ in\ water\)\(= v = 1600\ \text{ms}^{-1}\)
To Find
\(Distance\ from\ ship\ to\ seafloor = d = ?\)
Solution
By using the SONAR distance formula:
\[\begin{aligned} Distance &= \frac{Speed\ of\ Sound \times Time}{2} \\[6pt] d &= \frac{1600 \times 3.2}{2} \\[6pt] d &= \frac{5120}{2} \\[4pt] \boldsymbol{d} &= \boldsymbol{2560\ \text{m}} \end{aligned}\]
✅ \(d = 2560\ \text{m}\)
📐 Key Formulas – Sound
Wave Equation: \( v = f\lambda \)
Time Period: \( T = \frac{1}{f} \)
Frequency (chimes): \( f = \frac{n}{t} \)
SONAR / Echo Distance: \( d = \frac{v \times t}{2} \)
Distance (lightning): \( S = vt \)
Unit 10Thermal Physics
Unit 11Transfer of Thermal Energy
Unit 12Waves
Unit 13✨ Sound
Unit 14Light
Unit 15Electrostatics
Unit 16Electricity
Unit 17Electromagnetism
Unit 18Electromagnetic Induction & EM Waves
Unit 19Electronics
Unit 20Atomic and Nuclear Physics
Unit 21Space and Environment
📖 Complete syllabus coverage for Class 10 Physics (PECTAA 2026) – Units 10 to 21
💡 Exam Tip:
For numerical problems, always write Given Data, To Find, and step-by-step Solution with formulas. Pay attention to units and conversions. These problems follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus