Select the correct option for each of the following questions A boy standing in front of a wall produces two whistles per second. He notices that the sound of his whistling coincides with the echo. The echo is heard only once when whistling is stopped. Calculate the distance between the boy and the wall. (The speed of sound in air = 320 m/s)
If the speed of sound is increased by 16 ms-1 and the boy moves 4 m away from the wall then in how much time will he hear the echo of the first whistle? (a) 0.525 s (b) 0.5 s (c) 0.48 s (d) 0.3 s
Topic: Echo and the speed of sound
Try it
Pick an option and check your answer. The full worked answer is below either way.
Answer
(b) From the stem, the boy whistles twice per second and the echo coincides with the next whistle, so the echo returns in 0.5 s. At the original speed of 320 m s⁻¹: 2d = 320 × 0.5 = 160, so d = 80 m.
New conditions: new speed v′ = 320 + 16 = 336 m s⁻¹ new distance d′ = 80 + 4 = 84 m
For an echo, 2d′ = v′ × t: t = 2d′ ÷ v′ = (2 × 84) ÷ 336 = 168 ÷ 336 = 0.5 s
Answer: (b) 0.5 s
Both quantities change, so change both before substituting: the speed goes up by 16 to 336 m s⁻¹ and the distance goes up by 4 to 84 m.
It is a coincidence — and a neat one — that the time comes out unchanged at 0.5 s. The distance rose by 5 % and the speed rose by 5 %, so they cancel. Do not let that make you think the question was a trick; work it through and the 0.5 s falls out.
Keep the factor of 2: the sound goes to the wall and back.
From the stem, the boy whistles twice per second and the echo coincides with the next whistle, so the echo returns in 0.5 s. At the original speed of 320 m s⁻¹: 2d = 320 × 0.5 = 160, so d = 80 m.
New conditions: new speed v′ = 320 + 16 = 336 m s⁻¹ new distance d′ = 80 + 4 = 84 m
For an echo, 2d′ = v′ × t: t = 2d′ ÷ v′ = (2 × 84) ÷ 336 = 168 ÷ 336 = 0.5 s
Answer: (b) 0.5 s