Numerical Problems — Chapter 5: Acoustics
Important Formulas:
| Formula | Use |
|---|---|
| v = fλ | Speed = frequency × wavelength |
| T = 1/f | Time period = 1/frequency |
| d = vt | Distance = speed × time |
| d = vt/2 | SONAR depth formula (echo) |
| Minimum echo distance = 17 m | At v = 340 m/s, t = 0.1 s |
Useful values:
- Speed of sound in air: v ≈ 340 m/s (at room temperature)
- Speed of sound in water: v ≈ 1500 m/s
- Speed of sound in steel: v ≈ 5100 m/s
- Human hearing: 20 Hz – 20,000 Hz
Problem 1. A sound wave has a frequency of 500 Hz and a wavelength of 0.68 m. Find its speed.
- Frequency, f = 500 Hz
- Wavelength, λ = 0.68 m
- Speed, v = ?
v = fλ = 500 × 0.68 = 340 m/s
Problem 2. A sound wave travels with a speed of 340 m/s and has a frequency of 1700 Hz. Find its wavelength.
- Speed, v = 340 m/s
- Frequency, f = 1700 Hz
- Wavelength, λ = ?
v = fλ → λ = v/f = 340/1700 = 0.2 m = 20 cm
Problem 3. A person claps near a cliff and hears the echo after 2 seconds. If the speed of sound in air is 340 m/s, find the distance of the cliff.
- Time for echo, t = 2 s
- Speed, v = 340 m/s
- Distance to cliff, d = ?
Total distance = v × t = 340 × 2 = 680 m
Sound travels to cliff and back, so: d = 680/2 = 340 m
Problem 4. A SONAR device sends an ultrasound pulse and receives the echo after 6 seconds. If the speed of sound in sea water is 1500 m/s, find the depth of the ocean.
- Time for echo, t = 6 s
- Speed in water, v = 1500 m/s
- Depth, d = ?
d = (v × t) / 2 = (1500 × 6) / 2 = 9000 / 2 = 4500 m
Problem 5. A bat emits ultrasound of frequency 50,000 Hz. If the speed of sound in air is 340 m/s, find the wavelength of this ultrasound.
- Frequency, f = 50,000 Hz = 5 × 10⁴ Hz
- Speed, v = 340 m/s
- Wavelength, λ = ?
λ = v/f = 340 / 50,000 = 0.0068 m = 6.8 mm
Problem 6. A sound has a time period of 0.005 seconds. Find: (a) Frequency of the sound (b) Whether it is in the audible range
- Time period, T = 0.005 s
(a) Frequency:
f = 1/T = 1/0.005 = 200 Hz
(b) Audible range:
Audible range: 20 Hz – 20,000 Hz
200 Hz is between 20 Hz and 20,000 Hz → Yes, it is audible.
Problem 7. Two students A and B are standing on opposite sides of a 1000 m long corridor. Student A shouts and both students hear the echo from the far wall. Who hears the echo first? Find the time for each. (Speed of sound = 340 m/s; A is at one end, B is 200 m from A)
- Length of corridor = 1000 m
- A is at position 0 m; B is at position 200 m
- Speed, v = 340 m/s
For student A (at 0 m): Far wall is at 1000 m. Total distance = 1000 + 1000 = 2000 m (to wall and back to A) t_A = 2000/340 = 5.88 s
For student B (at 200 m): Far wall is at 1000 m → 800 m from B. Sound goes from A (0 m) to wall (1000 m) = 1000 m, then back 800 m to B = 1800 m total. t_B = 1800/340 = 5.29 s
Student B hears the echo first (t_B < t_A).
Problem 8. A tuning fork completes 256 vibrations per second. Find: (a) Its frequency (b) Its time period (c) If sound speed is 340 m/s, the wavelength of the sound produced
(a) Frequency: f = 256 vibrations/second = 256 Hz
(b) Time period: T = 1/f = 1/256 = 0.0039 s ≈ 3.9 × 10⁻³ s
(c) Wavelength: λ = v/f = 340/256 = 1.33 m
Summary: Key Relationships
| Relationship | Formula |
|---|---|
| Speed, frequency, wavelength | v = fλ |
| Frequency and time period | f = 1/T |
| Echo distance | d = vt/2 |
| Minimum echo distance (air) | 17 m (at v = 340 m/s, t_min = 0.1 s) |
📚 Samacheer Kalvi 10th Science Guide — Buy Online