Home / IB MYP Practice Questions and Resources / IB MYP 4-5 Physics- Echoes and reflection of sound- Study Notes

IB MYP 4-5 Physics- Echoes and reflection of sound- Study Notes

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

IB MYP 4-5 Physics Study Notes – All topics

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: \(\boxed{1800 \, \text{m}}\)

Scroll to Top