Sound – Numerical Problems with Stepwise Solutions

CBSE Class 8 Science – Chapter 13: Sound | Numerical Problems with Stepwise Solutions

CBSE Class 8 Science – Chapter Wise Study Materials Based on NCERT

Chapter 13: Sound – Numerical Problems with Stepwise Solutions

Class: 8

Subject: Science

Chapter: 13 – Sound

Board: CBSE

Suitable for: Annual Examination • Periodic Tests • Class Tests • Pre-Board / School Level Board Exam Pattern

These Numerical Problems with Stepwise Solutions are designed strictly as per the NCERT syllabus, making them ideal for CBSE Class 8 board exams standard.

Content Bank – Important Formulas of Chapter 13: Sound

Speed, Distance and Time: Speed (v) = Distance (s) ÷ Time (t)
Distance travelled by sound: Distance (s) = v × t
Frequency: Frequency (f) = Number of vibrations (n) ÷ Time (t)
Time Period: Time period (T) = Time (t) ÷ Number of vibrations (n)
Relation between Frequency and Time Period: f = 1 ÷ T and T = 1 ÷ f
Speed, Frequency and Wavelength: v = f × λ
Unit reminders: Speed in m s⁻¹, distance in m, time in s, frequency in Hz.
Approximate speed of sound in air (room temperature): v ≈ 340 m s⁻¹

Below are 20 topic-wise numerical problems from Chapter 13: Sound with full, step-by-step solutions. They focus on speed of sound, distance, time, frequency, time period and simple applications, for clear NCERT Class 8 level understanding.

Topic-wise Numerical Problems – Chapter 13: Sound

A. Basic Speed–Distance–Time Problems with Sound

Q1. The speed of sound in air is taken as 340 m s⁻¹. How long will sound take to travel a distance of 680 m?

Given: Speed of sound, v = 340 m s⁻¹; Distance, s = 680 m

To find: Time taken, t

Formula: v = s ÷ t → t = s ÷ v

Step 1: Substitute values:

t = 680 ÷ 340

Step 2: Divide:

t = 2 s

Answer: Sound will take 2 s to travel 680 m.

Q2. A boy standing at some distance from a wall shouts and hears the echo after 3 s. If the speed of sound in air is 330 m s⁻¹, calculate the distance between the boy and the wall.

Given: Total time for sound to travel to the wall and back, t = 3 s; Speed of sound, v = 330 m s⁻¹.

To find: Distance between boy and wall, d.

Concept: In echo, sound travels to the wall and back.

So total distance travelled by sound = 2d.

Step 1 (Use formula): v = s ÷ t, so s = v × t

s = 330 × 3 = 990 m

Step 2: 2d = 990 → d = 990 ÷ 2 = 495 m

Answer: The boy is at a distance of 495 m from the wall.

Q3. A school bell is 1700 m away from a playground. If the speed of sound in air is 340 m s⁻¹, how much time will it take for the sound of the bell to reach the students in the playground?

Given: s = 1700 m, v = 340 m s⁻¹

To find: t

Formula: t = s ÷ v

Step 1: Substitute values:

t = 1700 ÷ 340

Step 2: Divide:

t = 5 s

Answer: Sound of the bell will reach the students in 5 seconds.

Q4. An explosion occurs at a mine. A person standing 1020 m away hears the sound after 3 s. Find the speed of sound in air from this observation.

Given: Distance, s = 1020 m; Time, t = 3 s

To find: Speed of sound, v

Formula: v = s ÷ t

Step 1: Substitute values:

v = 1020 ÷ 3

Step 2: Divide:

v = 340 m s⁻¹

Answer: The speed of sound in air is 340 m s⁻¹.

B. Frequency, Time Period and Number of Vibrations

Q5. A vibrating body completes 150 vibrations in 3 s. Calculate its frequency.

Given: Number of vibrations, n = 150; Time, t = 3 s

To find: Frequency, f

Formula: f = n ÷ t

Step 1: Substitute values:

f = 150 ÷ 3

Step 2: Divide:

f = 50 Hz

Answer: The frequency of the vibrating body is 50 Hz.

Q6. A tuning fork has a frequency of 256 Hz. How many vibrations does it complete in 4 s?

Given: Frequency, f = 256 Hz; Time, t = 4 s

To find: Number of vibrations, n

Formula: f = n ÷ t → n = f × t

Step 1: Substitute values:

n = 256 × 4 = 1024

Answer: The tuning fork completes 1024 vibrations in 4 seconds.

Q7. A simple pendulum completes 40 vibrations in 20 s. Find (a) its frequency and (b) its time period.

Given: n = 40; t = 20 s

(a) Frequency:

f = n ÷ t = 40 ÷ 20 = 2 Hz

(b) Time period:

We can use T = 1 ÷ f

T = 1 ÷ 2 = 0.5 s

Answer: Frequency = 2 Hz; Time period = 0.5 s.

Q8. The time period of a vibrating body is 0.02 s. What is its frequency?

Given: Time period, T = 0.02 s

To find: Frequency, f

Formula: f = 1 ÷ T

Step 1: Substitute the value:

f = 1 ÷ 0.02

Step 2: Calculate:

f = 50 Hz

Answer: The frequency of the vibrating body is 50 Hz.

Q9. The frequency of a sound is 500 Hz. What is its time period?

Given: f = 500 Hz

To find: T

Formula: T = 1 ÷ f

Step 1: Substitute the value:

T = 1 ÷ 500

Step 2: Write answer in seconds:

T = 0.002 s

Answer: The time period of the sound is 0.002 s.

C. Speed, Frequency and Wavelength (v = f × λ)

Q10. A sound wave has a frequency of 400 Hz and travels in air with a speed of 320 m s⁻¹. Calculate its wavelength.

Given: f = 400 Hz; v = 320 m s⁻¹

To find: Wavelength, λ

Formula: v = f × λ → λ = v ÷ f

Step 1: Substitute values:

λ = 320 ÷ 400

Step 2: Calculate:

λ = 0.8 m

Answer: The wavelength of the sound wave is 0.8 m.

Q11. A sound wave travelling in air has a wavelength of 0.5 m. If the speed of sound is 340 m s⁻¹, find its frequency.

Given: λ = 0.5 m; v = 340 m s⁻¹

To find: f

Formula: v = f × λ → f = v ÷ λ

Step 1: Substitute:

f = 340 ÷ 0.5

Step 2: Calculate:

f = 680 Hz

Answer: The frequency of the sound wave is 680 Hz.

Q12. A sound wave of frequency 250 Hz has a wavelength of 1.2 m in a medium. Calculate the speed of sound in that medium.

Given: f = 250 Hz; λ = 1.2 m

To find: v

Formula: v = f × λ

Step 1: Substitute:

v = 250 × 1.2

Step 2: Multiply:

v = 300 m s⁻¹

Answer: The speed of sound in the medium is 300 m s⁻¹.

Q13. In a medium, sound travels at 330 m s⁻¹. If the wavelength of the sound is 0.6 m, what is its frequency and time period?

Given: v = 330 m s⁻¹; λ = 0.6 m

To find: f and T

Step 1 (frequency): Use f = v ÷ λ

f = 330 ÷ 0.6

f = 550 Hz

Step 2 (time period): Use T = 1 ÷ f

T = 1 ÷ 550 ≈ 0.00182 s

Answer: Frequency = 550 Hz; Time period ≈ 0.00182 s.

D. Application-Based Numerical Problems (Real-life Situations)

Q14. A lightning flash is seen and the thunder is heard 4 s later. If the speed of sound in air is 340 m s⁻¹, estimate how far the lightning occurred from the observer. (Ignore the time taken by light to reach us.)

Given: Time delay, t = 4 s; v = 340 m s⁻¹

To find: Distance to lightning, s

Formula: s = v × t

Step 1: Substitute:

s = 340 × 4 = 1360 m

Answer: The lightning occurred about 1360 m (1.36 km) away from the observer.

Q15. Two students are standing 1020 m apart. One of them fires a cracker and the other hears the sound after 3 s. Calculate the speed of sound from this observation. Is this value close to the commonly accepted speed of sound in air?

Given: Distance, s = 1020 m; Time, t = 3 s

To find: v

Formula: v = s ÷ t

Step 1: Substitute:

v = 1020 ÷ 3 = 340 m s⁻¹

Step 2 (comparison): The commonly accepted speed of sound in air is about 340 m s⁻¹ at room temperature.

Answer: Speed of sound = 340 m s⁻¹, which matches the commonly accepted value for air.

Q16. An observer hears the sound of a distant drum 2.5 s after seeing it being struck. If the speed of sound is 340 m s⁻¹, find the distance of the drum from the observer.

Given: t = 2.5 s; v = 340 m s⁻¹

To find: s

Formula: s = v × t

Step 1: Substitute:

s = 340 × 2.5

Step 2: Multiply:

s = 850 m

Answer: The drum is at a distance of 850 m from the observer.

Q17. During a school experiment, students calculate the speed of sound by hitting two wooden blocks and measuring the time taken for the sound to travel 680 m. If they measure the time as 2 s, what speed of sound do they obtain?

Given: s = 680 m; t = 2 s

To find: v

Formula: v = s ÷ t

Step 1: Substitute:

v = 680 ÷ 2 = 340 m s⁻¹

Answer: The students obtain the speed of sound as 340 m s⁻¹.

E. Mixed Conceptual Numerical Problems

Q18. A sound wave with frequency 300 Hz has a time period of 0.0033 s according to a student. Check whether the student’s value of time period is correct using the relation between frequency and time period.

Given: f = 300 Hz; student’s T = 0.0033 s

To check: Correct value of T using T = 1 ÷ f

Step 1: Calculate correct time period:

T = 1 ÷ 300

T ≈ 0.00333 s

Step 2: Compare values:

Student’s value = 0.0033 s, Correct value ≈ 0.00333 s.

They are very close (difference is very small), so the student’s value is approximately correct.

Answer: Correct T ≈ 0.00333 s; the student’s value 0.0033 s is acceptable at Class 8 level.

Q19. Sound travels in steel at about 5000 m s⁻¹. If a train is 2 km away on a steel track, how much time will the sound take to reach a person placing his ear on the track? (Take 1 km = 1000 m)

Given: v (steel) = 5000 m s⁻¹, Distance s = 2 km = 2000 m

To find: t

Formula: t = s ÷ v

Step 1: Substitute:

t = 2000 ÷ 5000

Step 2: Simplify fraction:

t = 0.4 s

Answer: Sound reaches the person through the track in 0.4 s.

Q20. A sound source produces waves of wavelength 0.85 m in air. If the speed of sound in air is 340 m s⁻¹, calculate (a) the frequency and (b) how many vibrations the source makes in 5 s.

Given: λ = 0.85 m; v = 340 m s⁻¹; time t = 5 s

(a) Frequency:

Use f = v ÷ λ

f = 340 ÷ 0.85

f = 400 Hz

(b) Number of vibrations in 5 s:

Use n = f × t

n = 400 × 5 = 2000

Answer: (a) Frequency = 400 Hz; (b) The source makes 2000 vibrations in 5 s.

These 20 numerical problems with detailed stepwise solutions cover core relations like speed, distance, time, frequency, time period and wavelength for Chapter 13: Sound, closely following the NCERT Class 8 Science syllabus and CBSE exam pattern.

Sound – Numerical Problems with Stepwise Solutions

CBSE Class 8 Science – Chapter Wise Study Materials Based on NCERT

Chapter 13: Sound – Numerical Problems with Stepwise Solutions

Content Bank – Important Formulas of Chapter 13: Sound

Topic-wise Numerical Problems – Chapter 13: Sound

A. Basic Speed–Distance–Time Problems with Sound

B. Frequency, Time Period and Number of Vibrations

C. Speed, Frequency and Wavelength (v = f × λ)

D. Application-Based Numerical Problems (Real-life Situations)

E. Mixed Conceptual Numerical Problems

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