On a bright winter's day, a coach notices an echo as he walks between a high school's playing field and the rear wall of its auditorium. He decides to estimate how far he stands from the reflective wall. He abruptly claps and determines that the sound pulse returns to his as an echo 0.523 seconds later (averaging several trials for enhanced reliability). The air is a frigid −10°C.Six months later, the coach decides to repeat his experiment on a hot summer day, standing at the same exact spot. This time, the echo takes a slightly shorter 0.484 seconds to return to the coach.(a)What is the temperature (in °C) of the air on the summer day? °C(b)How far (in m) does the coach stand from the reflective wall?
Question
On a bright winter's day, a coach notices an echo as he walks between a high school's playing field and the rear wall of its auditorium. He decides to estimate how far he stands from the reflective wall. He abruptly claps and determines that the sound pulse returns to his as an echo 0.523 seconds later (averaging several trials for enhanced reliability). The air is a frigid −10°C.Six months later, the coach decides to repeat his experiment on a hot summer day, standing at the same exact spot. This time, the echo takes a slightly shorter 0.484 seconds to return to the coach.(a)What is the temperature (in °C) of the air on the summer day? °C(b)How far (in m) does the coach stand from the reflective wall?
Solution
To solve this problem, we need to use the formula for the speed of sound in air, which is given by v = 331.4 + 0.6T, where v is the speed of sound in meters per second and T is the temperature in degrees Celsius.
(a) First, we need to find the speed of sound on the winter day. We know that the echo took 0.523 seconds to travel to the wall and back, so the time it took to travel one way is 0.523/2 = 0.2615 seconds. The speed of sound is the distance divided by the time, so v = d/t. We don't know the distance yet, but we can express it in terms of v: d = vt. So the speed of sound on the winter day is v = d/t = d/0.2615.
We also know that the temperature on the winter day was -10°C, so we can use the formula for the speed of sound to express v in terms of T: v = 331.4 + 0.6T = 331.4 + 0.6(-10) = 325.4 m/s.
Now we can set the two expressions for v equal to each other and solve for d: d/0.2615 = 325.4, so d = 0.2615 * 325.4 = 85.1 meters.
Next, we need to find the speed of sound on the summer day. We know that the echo took 0.484 seconds to travel to the wall and back, so the time it took to travel one way is 0.484/2 = 0.242 seconds. Using the same reasoning as before, the speed of sound on the summer day is v = d/t = d/0.242.
We can set this equal to the formula for the speed of sound in terms of T and solve for T: d/0.242 = 331.4 + 0.6T. Substituting the value we found for d, we get 85.1/0.242 = 331.4 + 0.6T. Solving for T, we get T = (85.1/0.242 - 331.4)/0.6 = 20.8°C.
(b) We already found the distance from the coach to the wall in part (a): it's 85.1 meters.
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