Sound – Numerical Problems with Stepwise Solutions
CBSE Class 9
Physics — Chapter 12: Sound
20 Numerical Problems with Stepwise Solutions — NCERT-aligned for CBSE Class 9 board exams
Content Bank — Important Formulas
Wave relation:
v = f × λ (v: speed in m·s−1, f: frequency in Hz, λ: wavelength in m)Speed of sound in air (approx.):
v ≈ 331 + 0.6 T m·s−1 (T in °C)Echo distance:
d = (v × Δt) / 2 (Δt = round-trip time)Beat frequency:
f_{beat} = |f_1 − f_2|Intensity vs amplitude:
I ∝ A^2Sonar / depth:
depth = (c × Δt) / 2 where c is speed in water.Problem 1
Find the wavelength of a sound wave of frequency
1200 Hz travelling in air at speed 343 m·s−1.- Use
λ = v / f. - Substitute:
λ = 343 / 1200m.
Calculation:
λ = 343 ÷ 1200 = 0.285833... m ≈ 0.286 m.Problem 2
A sound has wavelength
0.5 m in air where speed is 340 m·s−1. Calculate its frequency.- Use
f = v / λ. - Substitute:
f = 340 / 0.5Hz.
Calculation:
f = 340 ÷ 0.5 = 680 Hz.Problem 3
Estimate the speed of sound in air at
25 °C using the approximation v ≈ 331 + 0.6 T.- Substitute T = 25:
v ≈ 331 + 0.6 × 25.
Calculation:
v ≈ 331 + 15 = 346 m·s−1.Problem 4
An echo is heard 0.20 s after the original sound is produced. If speed of sound is
340 m·s−1, what is the distance to the reflecting surface?- Round-trip distance =
v × Δt = 340 × 0.20 = 68 m. - One-way distance = half of round-trip =
68 / 2 = 34 m.
Distance to reflector ≈ 34 m.
Problem 5
How long does it take for a sound to travel 100 m in air if speed is
343 m·s−1?- Use
time = distance / speed⇒t = 100 / 343s.
Calculation:
t = 100 ÷ 343 ≈ 0.291545 s ≈ 0.292 s.Problem 6
In a closed resonance tube the first resonance (fundamental) is observed at length
L = 0.167 m with tuning fork frequency 512 Hz. Find wavelength and the speed of sound from these observations (closed tube: λ = 4L for fundamental).- Wavelength:
λ = 4 L = 4 × 0.167 = 0.668 m. - Speed:
v = f × λ = 512 × 0.668m·s−1.
Calculation:
v = 512 × 0.668 = 342.016 m·s−1 ≈ 342.0 m·s−1.Problem 7
Two tuning forks of frequencies
440 Hz and 444 Hz are struck together. Calculate the beat frequency.- Use
f_{beat} = |f_1 − f_2| = |444 − 440|Hz.
Beat frequency = 4 Hz (loudness varies 4 times per second).
Problem 8
A research vessel emits a sonar ping and receives the echo after
3.0 s. If speed of sound in seawater is 1500 m·s−1, estimate the depth of the seafloor.- Round-trip distance =
c × Δt = 1500 × 3.0 = 4500 m. - Depth = half of round-trip =
4500 / 2 = 2250 m.
Estimated depth ≈ 2250 m.
Problem 9
If the amplitude of a sound wave doubles, by what factor does intensity change? (Given
I ∝ A2.)- If A → 2A, intensity I ∝ (2A)2 = 4 A2. So intensity increases by factor 4.
Intensity becomes 4 times the original.
Problem 10
Find the decibel level of intensity
I = 1.0 × 10−12 W·m−2 relative to reference I0 = 10−12 W·m−2. (Use L = 10 log10(I/I0) dB.)- Compute ratio
I/I0 = 1.0 × 10−12 / 1.0 × 10−12 = 1. - So
L = 10 log10(1) = 0 dB.
Decibel level = 0 dB (threshold of hearing reference).
Problem 11
How long does sound take to travel 2.0 km through steel if speed in steel is approx.
5000 m·s−1?- Distance = 2000 m. Time = distance / speed =
2000 / 5000 = 0.4 s.
Time ≈ 0.40 s.
Problem 12
Calculate the wavelength of a 256 Hz tone in air where v = 340 m·s−1.
- λ = v / f = 340 / 256.
Calculation:
λ = 340 ÷ 256 = 1.328125 m ≈ 1.33 m.Problem 13
If a clap produces an echo after 0.15 s (round-trip) and v = 340 m·s−1, how far is the reflecting wall from the clapper?
- Round-trip distance =
340 × 0.15 = 51 m. - One-way distance =
51 / 2 = 25.5 m.
Distance to wall ≈ 25.5 m.
Problem 14
In a measurement, wavelength has 2% uncertainty and frequency has 1% uncertainty. Estimate the percentage uncertainty in calculated speed
v = f × λ.- Relative errors add for multiplication: total ≈ 2% + 1% = 3%.
Percentage uncertainty in v ≈ 3%.
Problem 15
A source emits sound at frequency
500 Hz. If the source moves towards a stationary observer at 20 m·s−1 and v = 343 m·s−1, estimate the observed frequency (use the source-moving formula qualitatively: f' = f × v / (v − v_s)).- Substitute:
f' = 500 × 343 / (343 − 20).
Calculation:
f' ≈ 500 × 343 / 323 ≈ 530.96 Hz ≈ 531.0 Hz (apparent pitch increases).Problem 16
If a listener moves twice as far from a point source, by what factor does the intensity change approximately (inverse-square law)?
- Intensity I ∝ 1/r2. Doubling r → I scales by 1/(22) = 1/4.
Intensity becomes one quarter (0.25×) of original.
Problem 17
Find speed of sound when frequency is
400 Hz and wavelength is 0.85 m.- Use
v = f × λ = 400 × 0.85.
Calculation:
v = 400 × 0.85 = 340 m·s−1.Problem 18
A signal pulse travels 102 m (round-trip) in 0.30 s. Verify the speed of sound used and find the one-way distance.
- Speed v = distance / time = 102 / 0.30 = 340 m·s−1.
- One-way distance = 102 / 2 = 51 m.
Computed speed = 340 m·s−1; one-way distance = 51 m.
Problem 19
Why do low-frequency sounds travel farther outdoors than high-frequency sounds? Provide a short quantitative argument referring to absorption/scattering.
- Low-frequency sounds have longer wavelengths and suffer less atmospheric absorption and scattering by small obstacles; absorption roughly increases with frequency.
- Qualitatively, attenuation α often increases with frequency, so over distance d, intensity ∝ e−αd decays faster for higher α (higher f).
Thus low-frequency components (e.g., bass) are less attenuated and diffract around obstacles more, enabling them to be heard at greater distances.
Problem 20
What is the minimum one-way distance to a reflecting surface required to hear a distinct echo if the speed of sound is
340 m·s−1 and the ear requires at least 0.1 s delay?- Round-trip distance for 0.1 s =
340 × 0.1 = 34 m. - So minimum one-way distance =
34 / 2 = 17 m.
Minimum distance ≈ 17 m.