(a)(i) What is an echo? (ii) State two useful applications of echoes. (iii) Why are the walls, floors and ceilings of a recording studio heavily padded? (b)...

Assessment: WAEC SSCE - Physics - 2001 (Essay) Subject: Physics

Question 1 Report

(a)(i) What is an echo? (ii) State two useful applications of echoes.

(iii) Why are the walls, floors and ceilings of a recording studio heavily padded?

(b)(i) Explain timbre and overtones.

(ii) What is resonance?

(c) As a ship approaches a cliff, its siren is sounded and the echo is heard in the ship after 12 seconds. 2.1 minutes later the siren was sounded again and the echo was heard 8 seconds later. If the speed of sound in air is 340 ms\(^{-1}\), calculate the velocity at which the ship was approaching the cliff.

Answer Details

(a)(i) An echo is a reflected sound wave heard distinctly after the original sound, produced when sound is reflected from a hard, distant surface back to the listener.

(a)(ii) Two useful applications:

  • Depth sounding (echo sounding/SONAR) to measure the depth of the sea or locate shoals of fish and submarines.
  • Detecting flaws in metals and, in medicine, ultrasonic scanning of the body.

(a)(iii) The walls, floors and ceilings of a recording studio are heavily padded with soft, porous material to absorb sound and prevent reflection (echoes and reverberation), so that only clear, direct sound is recorded.

(b)(i) Timbre (quality) is the property of a musical note that lets the ear distinguish two notes of the same pitch and loudness from different sources; it depends on the number and relative strength of the overtones present. Overtones are the higher frequencies (above the fundamental) that accompany a note.

(b)(ii) Resonance is the condition in which a body is set into vibration with large amplitude by a periodic force whose frequency equals the natural frequency of the body.

(c) Velocity of the ship

Distance of the cliff at the first sounding \(= \tfrac{1}{2}(340)(12) = 2040\,\text{m}\).

Distance at the second sounding \(= \tfrac{1}{2}(340)(8) = 1360\,\text{m}\).

Distance travelled by the ship between the two soundings:

\[ 2040 - 1360 = 680\,\text{m} \]

Time between soundings \(= 2.1\,\text{min} = 126\,\text{s}\). Hence:

\[ v = \frac{680}{126} \approx 5.4\,\text{ms}^{-1} \]

The ship approaches the cliff at about \(5.4\,\text{ms}^{-1}\).

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