IB MYP 4-5 Physics- Echoes and reflection of sound- Study Notes - New Syllabus
IB MYP 4-5 Physics-Echoes and reflection of sound- Study Notes
Key Concepts
- Echoes and reflection of sound
Echoes and reflection of sound
Reflection of Sound
Sound waves, like light w+aves, can be reflected when they hit a large, hard surface (e.g., wall, building, cliff).
- The laws of reflection apply to sound as well:
- The angle of incidence = the angle of reflection.
- Incident wave, reflected wave, and normal all lie in the same plane.
- Applications:
- Megaphones and horns use reflection to direct sound in one direction.
- Concert halls use reflectors to distribute sound evenly.
- Stethoscopes use reflection of sound in tubes.
Echo
An echo is a reflected sound heard after the original sound when it bounces off a large surface such as a wall, mountain, or building.
- For an echo to be heard clearly:
- The reflecting surface must be at least 17 m away (because the human ear can only distinguish two sounds if the time gap is at least 0.1 s).
- Speed of sound in air ≈ 340 m/s → in 0.1 s, sound travels about 34 m → so the wall must be half that distance (17 m away).
- Formula for echo method:
\( v = \dfrac{2d}{t} \)
- \(v\) = speed of sound (m/s)
- \(d\) = distance to wall (m)
- \(t\) = time taken for echo (s)
Applications of Echoes
- Measuring speed of sound (echo method).
- SONAR (Sound Navigation and Ranging) used by ships and submarines to detect depth of sea and obstacles.
- Ultrasound scanning in medicine to see internal organs.
- Animals like bats and dolphins use echoes for navigation (echolocation).
Example:
A boy shouts near a mountain and hears the echo after 0.2 s. If the speed of sound is 340 m/s, calculate the distance of the mountain.
▶️ Answer/Explanation
Formula: \(d = \dfrac{vt}{2}\)
\(d = \dfrac{340 \times 0.2}{2} = 34 \, \text{m}\)
Final Answer: \(\boxed{34 \, \text{m}}\)
Example:
A person is 85 m away from a wall. How much time will it take for him to hear his echo if the speed of sound is 340 m/s?
▶️ Answer/Explanation
Total distance = \(2d = 2 \times 85 = 170 \, \text{m}\)
\(t = \dfrac{2d}{v} = \dfrac{170}{340} = 0.5 \, \text{s}\)
Final Answer: \(\boxed{0.5 \, \text{s}}\)
Example:
A ship sends a SONAR signal towards the seabed and receives the echo after 2.4 seconds. If the speed of sound in seawater is 1500 m/s, calculate the depth of the sea.
▶️ Answer/Explanation
Total distance traveled = \(v \times t = 1500 \times 2.4 = 3600 \, \text{m}\)
Depth = \(\dfrac{3600}{2} = 1800 \, \text{m}\)
Final Answer: \(\boxed{1800 \, \text{m}}\)