What are the first, second, and third frequencies audible from a 20 cm long organ pipe when (A) only one end is open and when (B
) both ends are open. The speed of sound through the air inside the organ pipe is 343m/s. (Hint: if a harmonic does not exist, it would not be heard.)
1 answer:
A) The answers are:
the first frequency - 428.75 Hz
the second frequency - 1286.25 Hz
the third frequency - 2143.75 Hz
The frequency (when the pipe is closed) is: f = v(2n - 1)/4L
v - the speed of sound
n - the frequency order
L - the length of the organ pipe
We know:
v = 343 m/s
L = 20 cm = 0.2 m
1. The first frequency (n = 1):
f = 343 * (2 * 1 - 1) / 4 * 0.2 = 343 * 1 / 0.8 = 428.75 Hz
2. The second frequency (n = 2):
f = 343 * (2 * 2 - 1) / 4 * 0.2 = 343 * 3 / 0.8 = 1286.25 Hz
3. The third frequency (n = 3):
f = 343 * (2 * 3 - 1) / 4 * 0.2 = 343 * 5 / 0.8 = 2143.75 Hz
B) The answers are:
the first frequency - 857.5 Hz
the second frequency - 1715 Hz
the third frequency - 2572.5 Hz
The frequency (when the pipe is open) is: f = vn/2L
v - the speed of sound
n - the frequency order
L - the length of the organ pipe
We know:
v = 343 m/s
L = 20 cm = 0.2 m
1. The first frequency (n = 1):
f = 343 * 1 / 2 * 0.2 = 343 / 0.4 = 857.5 Hz
2. The second frequency (n = 2):
f = 343 * 2 / 2 * 0.2 = 686 / 0.4 = 1715 Hz
3. The third frequency (n = 3):
f = 343 * 3 / 2 * 0.2 = 1029 / 0.4 = 2572.5 Hz
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It is B or 2. Hope this helps
Answer:
Priya's age = 15
Amirah's age = 19
Shirley's age = 13
Step-by-step explanation:
Let x be the age of Priya.
According to the question,
Amirah's age = x + 4
Shirley's age = x - 2
Their sum is 47.
So,
x + 4 + x + x - 2 = 47
3x + 2 = 47
3x = 47 - 2
3x = 45
x = 45/3
x = 15
So,
Priya's age = 15
Amirah's age = 15 + 4 = 19
Shirley's age = 15 - 2 = 13
Answer:
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Step-by-step explanation:
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40/5 = 8
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The second one is correct