Sound — Practice

Question 1

What kind of wave is sound — transverse or longitudinal? Justify your answer.

Accepted Answer

Step 1 — Recall the distinction.
  Transverse: particles vibrate PERPENDICULAR to wave propagation.
  Longitudinal: particles vibrate PARALLEL to wave propagation.

Step 2 — Sound mechanism.
  Sound in a gas: air particles vibrate back-and-forth in the same direction as the wave moves (creating compressions and rarefactions).

✦ Answer: Sound is a LONGITUDINAL wave. Particles in the medium oscillate parallel to the direction of wave propagation, alternately compressing and rarefying the medium.

Question 2

A wave has frequency 500 Hz and wavelength 0.5 m. Find its speed.

Accepted Answer

Step 1 — Use v = fλ.

Step 2 — v = 500 × 0.5 = 250 m/s.

✦ Answer: 250 m/s.

Question 3

Differentiate pitch and loudness of sound. On which wave property does each depend?

Accepted Answer

Step 1 — Pitch.
  • Perceived 'highness' or 'lowness' of sound.
  • Depends on FREQUENCY.
  • Higher frequency → higher pitch. A whistle (high f) is shrill; a tuba (low f) is bassy.

Step 2 — Loudness.
  • Perceived intensity / volume.
  • Depends on AMPLITUDE.
  • Higher amplitude → louder. A drum can be played softly or loudly at the SAME pitch.

✦ Answer: Pitch is determined by FREQUENCY (Hz). Loudness is determined by AMPLITUDE (the maximum displacement). They are independent — same loudness can have different pitches, and vice versa.

Question 4

Classify each as audible, infrasonic, or ultrasonic for humans: (a) 10 Hz, (b) 500 Hz, (c) 25,000 Hz, (d) 18,000 Hz.

Accepted Answer

Step 1 — Recall the human range.
  Audible: 20 Hz to 20,000 Hz (20 kHz).
  Below 20 Hz: INFRASONIC.
  Above 20 kHz: ULTRASONIC.

Step 2 — Classify each.
  (a) 10 Hz: below 20 Hz → INFRASONIC.
  (b) 500 Hz: in 20 Hz to 20 kHz → AUDIBLE.
  (c) 25,000 Hz: above 20 kHz → ULTRASONIC.
  (d) 18,000 Hz: below 20 kHz → AUDIBLE.

✦ Answer: (a) Infrasonic, (b) Audible, (c) Ultrasonic, (d) Audible (high-pitched but still audible).

Question 5

Why do we hear an echo only when we are at least 17 m away from a wall?

Accepted Answer

Step 1 — Recall: for a distinct echo, reflected sound must reach our ears at least 0.1 s after the original sound (otherwise our brain merges them).

Step 2 — In 0.1 s, sound travels: 340 × 0.1 = 34 m.

Step 3 — This is the ROUND TRIP distance (to the wall and back). The wall is at half this: 34/2 = 17 m.

✦ Answer: At 17 m, the round trip is 34 m, taking 0.1 s — the minimum gap for our brain to distinguish original from reflected. Closer than 17 m, the two merge as reverberation; farther than 17 m, distinct echo.

Question 6

A radio station broadcasts at frequency 100 MHz. Find the wavelength. (Speed of EM waves in air = 3 × 10⁸ m/s)

Accepted Answer

Step 1 — Convert MHz to Hz.
  100 MHz = 100 × 10⁶ Hz = 10⁸ Hz.

Step 2 — Use v = fλ ⇒ λ = v/f.
  λ = (3 × 10⁸) / 10⁸ = 3 m.

✦ Answer: Wavelength = 3 m.

(Note: this example uses light/radio speed, not sound speed. Same v = fλ formula applies to all waves.)

Question 7

A SONAR device sends an ultrasonic pulse into the sea. The reflected pulse from the ocean floor is received 4 s later. If sound speed in sea water is 1500 m/s, find the depth.

Accepted Answer

Step 1 — Apply SONAR formula.
  d = (v × t) / 2.

Step 2 — Substitute.
  d = (1500 × 4) / 2 = 6000 / 2 = 3000 m.

✦ Answer: Depth = 3000 m = 3 km.

Reminder: divide by 2 because the pulse travels DOWN and BACK — the time t covers the round trip.

Question 8

A tuning fork produces 256 vibrations per second. If sound travels at 340 m/s in air, find the wavelength of the sound.

Accepted Answer

Step 1 — Identify givens.
  Frequency f = 256 Hz. Speed v = 340 m/s.

Step 2 — Use v = fλ ⇒ λ = v/f.
  λ = 340 / 256 ≈ 1.33 m.

✦ Answer: Wavelength ≈ 1.33 m.

Question 9

A man fires a gun and hears the echo from a cliff 2 s later. If the speed of sound is 340 m/s, find the distance of the cliff from the man.

Accepted Answer

Step 1 — Use d = v × t / 2.

Step 2 — Substitute.
  d = (340 × 2) / 2 = 340 m.

✦ Answer: Distance to the cliff = 340 m.

Why divide by 2: the sound travels TO the cliff and BACK. 2 s is the round-trip time; the one-way distance is half.

Question 10

Why does sound travel faster in iron than in air?

Accepted Answer

Step 1 — Recall: speed of sound depends on the elasticity and density of the medium.
  v = √(B / ρ) for fluids, where B is the bulk modulus (a measure of stiffness) and ρ is density.
  For solids: v = √(E / ρ), where E is Young's modulus (rigidity).

Step 2 — Compare iron and air.
  • Iron is FAR stiffer (Young's modulus ~10¹¹ Pa) than air (bulk modulus ~10⁵ Pa).
  • Iron is also denser, but the elasticity dominates.
  • The ratio Y/ρ for iron >> B/ρ for air.

Step 3 — Result.
  Sound in iron ≈ 5960 m/s. In air ≈ 343 m/s. About 17× faster.

✦ Answer: Iron's particles are tightly bound by very strong inter-atomic forces — they transmit vibrations far more efficiently than the loosely-connected air molecules. The stiffness (elasticity) of iron vastly exceeds air, so vibrations propagate much faster.

Question 11

A boy stands between two parallel cliffs and claps his hands. He hears two distinct echoes after 2 s and 4 s respectively. Find the distance between the two cliffs. (Speed of sound = 340 m/s)

Accepted Answer

Step 1 — Distance to closer cliff.
  Round-trip 2 s at 340 m/s = 680 m.
  One-way distance = 680/2 = 340 m.

Step 2 — Distance to farther cliff.
  Round-trip 4 s at 340 m/s = 1360 m.
  One-way distance = 1360/2 = 680 m.

Step 3 — Total distance between cliffs.
  340 + 680 = 1020 m.

✦ Answer: Distance between the two cliffs = 1020 m.

Question 12

A wave has time period 0.005 s and travels at 300 m/s in a medium. Find its frequency and wavelength.

Accepted Answer

Step 1 — Frequency from time period.
  f = 1/T = 1/0.005 = 200 Hz.

Step 2 — Wavelength from v = fλ.
  λ = v/f = 300/200 = 1.5 m.

✦ Answer: Frequency = 200 Hz; Wavelength = 1.5 m.

Question 13

A sound wave has frequency 500 Hz. Find its wavelength in (a) air, (b) water, (c) steel. (Speeds: air 340 m/s, water 1500 m/s, steel 5960 m/s).

Accepted Answer

Step 1 — Frequency is the SAME in any medium.
  Wavelength changes because v changes. Use λ = v/f.

Step 2 — Compute each.
  (a) Air: λ = 340 / 500 = 0.68 m = 68 cm.
  (b) Water: λ = 1500 / 500 = 3.0 m.
  (c) Steel: λ = 5960 / 500 = 11.92 m.

Step 3 — Interpret.
  Same 'note' (same frequency, same pitch) has very different wavelengths in different media.
  But the EAR detects frequency (which stays constant) — so the note sounds the same to you whether the sound travels through air or water.

✦ Answer: (a) 0.68 m in air, (b) 3.0 m in water, (c) 11.92 m in steel. Frequency conserved; wavelength scales with the medium's sound speed.

Question 14

Bats use ultrasonic waves for navigation. Explain how a bat can find its prey in the dark.

Accepted Answer

Step 1 — Bat emits ultrasonic clicks.
  Frequency 20 kHz to 120 kHz (mostly ultrasonic; some clicks audible to humans as the bat exhales). Short pulses, several per second.

Step 2 — Pulses reflect off objects.
  When a click hits a flying insect or a tree branch, it reflects back. The bat's specialized ears detect the echo.

Step 3 — Bat measures the time delay.
  Distance = (v × t)/2 (same SONAR formula). With ultrasound at 340 m/s in air, and time resolution of microseconds, the bat can locate a mosquito-sized insect within centimetres.

Step 4 — Bat measures the Doppler shift.
  If the insect is moving toward the bat, the echo's frequency is slightly higher than the emitted click (Doppler effect). The bat reads this to estimate the prey's velocity. (Beyond Class 9, but a useful concept.)

Step 5 — Bat builds an internal 3D 'map' from clicks and echoes.
  Many pulses per second + slight angular separation between the two ears = a continuous 3D sound image. The bat can fly through dense forest in pitch dark.

✦ Answer: Bats emit ultrasonic clicks, listen for echoes, and use time-of-flight + intensity + frequency-shift to triangulate prey position and velocity in 3D — a biological version of SONAR (called ECHOLOCATION). Insects (especially moths) have evolved counter-measures (ultrasonic listening, evasive turns) — a continuous evolutionary arms race.

Question 15

If sound takes 6 s to travel through 2 km of water and 1.0 s in air for 340 m, calculate the ratio of speeds of sound in water and air.

Accepted Answer

Step 1 — Speed in water.
  v_water = distance / time = 2000 m / 6 s ≈ 333 m/s.

Wait, this seems too slow. Let me reread...

Reinterpret: 'takes 6 s' might be a typo for the wrong number. Let me use what's given.

Step 1 — Per the given numbers.
  v_water = 2000/6 ≈ 333 m/s.
  v_air = 340/1.0 = 340 m/s.

Step 2 — Ratio.
  v_water / v_air = 333 / 340 ≈ 0.98.

Step 3 — Note.
  In reality, v_water ≈ 1500 m/s and v_air ≈ 340 m/s, so v_water / v_air ≈ 4.4. The numbers in this problem appear to use different data — possibly distance in water was meant to be 8 km. Let me solve with both interpretations.

Likely intended: 8 km in 6 s → v_water = 8000/6 ≈ 1333 m/s. Ratio = 1333/340 ≈ 3.9. Closer to the expected ratio of ~4.

✦ Answer (using given numbers): v_water = 333 m/s, v_air = 340 m/s. Ratio ≈ 0.98 (close to 1).
  Likely intended answer (8 km, 6 s): v_water = 1333 m/s, v_air = 340 m/s. Ratio ≈ 3.9.

Note: Always check whether the given numbers are physically plausible.

Question 16

Give two medical and two industrial applications of ultrasound.

Accepted Answer

Step 1 — Medical applications.

(i) Diagnostic imaging — Pregnancy scans, echocardiograms, organ examinations. Ultrasound penetrates soft tissue, reflects at boundaries (e.g., between fluid and solid), and is non-ionising (unlike X-rays, so it's safe for pregnant women).

(ii) Lithotripsy — Treating kidney stones by focusing high-intensity ultrasound to break stones into small pieces that can pass naturally.

Step 2 — Industrial applications.

(iii) Ultrasonic cleaning — Jewellery, surgical instruments, electronics components are cleaned in ultrasonic baths. Tiny bubbles (cavitation) created by the ultrasound scrub away dirt at microscopic level.

(iv) Non-destructive testing — Detecting cracks in metal parts (aircraft, pipelines, bridges) using ultrasonic pulses. Faults show up as unexpected reflections.

(Bonus: Echolocation in bats and dolphins; ultrasonic welding of plastics; SONAR for fisheries and sea floor mapping.)

✦ Answer: Medical — diagnostic imaging (pregnancy/cardiac scans), lithotripsy for kidney stones. Industrial — ultrasonic cleaning of jewellery/instruments, non-destructive testing for cracks in metal.

Practice — Sound

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