Maria sings near the top of a mountain. She directs her voice towards the face o
ID: 1607486 • Letter: M
Question
Maria sings near the top of a mountain. She directs her voice towards the face of another, taller mountain nearby. A valley lies between two mountains.
1) If the temperature of the outside air is -1.5 degrees celsius, what is the speed of the sound?
2) If she is singing an "A" note (440 Hz), what is the wavelength of this wave?
3) If she hears an echo after 0.75 seconds, how far away is the other mountain
4) Suppose Maria and her friend Susan both sing "A" notes towards a third friend, Ramon. Suppose Maria is 3 meters from Raon, and Susan is 6 meters from Ramon. Compared to just one of them singing, will Ramon hear a louder sound, a softer sound or the same volume of sound?
5) Same question as 4- but now maria is 3 meters and Susan is 6.375 meters
Explanation / Answer
speed of sound v = 331.4 + (0.6*t)
1)
speed of sound v = 331.4 - (0.6*1.5) = 330.5 m/s
(2)
wavelength ,lambda = v/f = 330.5/440 = 0.75 m
(3)
distance travelled by echo = 2d
time taken t = 0.75 s
speed v = 2d/t
330.5 = 2*d/0.75
distance between two mountains d = 124 m
-----------------
(4)
path difference dr = r2 - r1 = 6 - 3 = 3m
to hear louder the waves interfere constructively
for louder path difference = n*lambda
n = 1 ,2 , 3, 4 ....
n = 3/0.75 = 4
Raon hears a louder volume
(5)
path difference = 6.375 - 3 = 3.375 m
for softer path difference dr = n*lambda/2
n = 1 , 3, 5, 7 , 9 , 11
n*0.75/2 = 3.375
n = 9
Ramon hears softer sound