8 min readConcept sheetPart II (2 marks) on the laws or echo; Part III (4 marks) and Part IV (7 marks) echo numericals

Reflection of sound and echo

Sound reflects like light: the angle of incidence equals the angle of reflection. An echo is reflected sound heard separately; it needs a reflector at least 17.2 m away.

Laws∠i = ∠r; incident sound, reflected sound and normal in one plane

Persistence of hearingAbout 0.1 s

Minimum distance for an echo17.2 m (at 344 m/s)

Echo formulad = vt/2

Laws of reflection of sound

Sound bounces off hard surfaces like walls, cliffs and buildings. This is reflection of sound.
First law: the angle of incidence equals the angle of reflection (∠i = ∠r). Both are measured from the normal, the line at right angles to the surface.
Second law: the incident sound, the reflected sound and the normal at the point of incidence all lie in the same plane.
Hard, smooth surfaces reflect sound well. Soft surfaces like curtains, cushions and carpets absorb it.

How sound reflects at different surfaces

At a denser mediumFor example, sound in air hitting a wall. A compression comes back as a compression, and a rarefaction as a rarefaction.

At a rarer mediumFor example, sound in water reaching the surface with air. A compression comes back as a rarefaction, and a rarefaction as a compression.

At a concave surfaceA surface curved inward. The reflected sound converges (comes together) at a point.

At a convex surfaceA surface curved outward. The reflected sound diverges (spreads out).

Echo and persistence of hearing

An echo is the sound heard again after reflection from a distant surface. Our ear keeps the feeling of a sound for about 0.1 s after it stops. This is persistence of hearing. To be heard separately, the echo must reach us at least 0.1 s after the original sound. Conditions for an echo: (i) the reflecting surface must be at least 17.2 m away. (ii) It must be large and hard. (iii) The sound must be loud enough to be heard after reflection.

Worked example: minimum distance for an echo

Problem

Show that the minimum distance needed to hear an echo is 17.2 m, when the speed of sound is 344 m/s.

Working

Let d be the distance to the reflector. Sound goes to it and comes back, so the total path is 2d.
Speed = distance ÷ time, so v = 2d/t, giving d = vt/2.
For a separate echo, t must be at least 0.1 s.
d = (344 m/s × 0.1 s) ÷ 2 = 34.4 m ÷ 2 = 17.2 m.

Answer

Minimum distance = 17.2 m

Worked example: echo from a hill and the speed of sound

Problem

(a) A boy shouts near a hill at Kodaikanal and hears the echo 2 s later. The speed of sound is 340 m/s. How far away is the hill? (b) An echo from a wall 85 m away returns in 0.5 s. Find the speed of sound.

Working

(a) Formula: d = vt/2.
d = (340 m/s × 2 s) ÷ 2 = 680 ÷ 2 = 340 m.
(b) Formula: v = 2d/t.
v = (2 × 85 m) ÷ 0.5 s = 170 ÷ 0.5 = 340 m/s.

Answer

(a) 340 m (b) 340 m/s

In real life

Shout inside an empty temple mandapam or near a rocky hill and you may hear an echo. Cinema halls cover their walls with soft panels and curtains to absorb sound, so echoes do not blur the dialogue.

Exam angle

2 marks: “State the laws of reflection of sound.” (i) The angle of incidence equals the angle of reflection. (ii) The incident sound, the reflected sound and the normal lie in the same plane.

2 marks: “What is an echo? Why can't we hear an echo in a small room?” Sound heard again after reflection. In a small room the walls are closer than 17.2 m, so the reflection returns within 0.1 s and merges with the original sound.

7 marks: “What is an echo? State the conditions for an echo, derive the minimum distance needed and solve a problem.” Definition; persistence of hearing 0.1 s; d = vt/2 = (344 × 0.1) ÷ 2 = 17.2 m; large, hard reflector; example: echo after 2 s at 340 m/s means the reflector is 340 m away.

Remember it

Echo needs 17.2: 0.1 s of hearing memory × 344 m/s, then ÷ 2 for the round trip.