1) The lowest tone to resonate in an open pipe of length L is 1600 Hz. What is the frequency of the lowest tone that will resonate in a closed pipe of length 2L?
a) 100 Hz
b) 1.60 Hz
c) 200 Hz
d) 0.800 kHz
2) Which of the following does not increase as a sound becomes louder?
a)frequency
b)velocity
c)period
d) wavelength
e) amplitude
3) The lowest tone to resonate in an open pipe of length L is 300 Hz. Which of the following frequencies will not resonate in the same pipe?
a) 800 Hz
b) 600 Hz
c) 400 Hz
d) 900 Hz
1. d
2. d
3. c
1) The frequency of the lowest tone that will resonate in a closed pipe of length 2L can be found using the formula:
f = (2n - 1) v / 4L
where f is the frequency, n is the harmonic number (1 for the fundamental frequency), v is the velocity of sound (assumed to be constant), and L is the length of the pipe.
In this case, the lowest tone to resonate in an open pipe of length L is 1600 Hz. To find the frequency of the lowest tone in a closed pipe of length 2L, we need to find n such that the frequency is the lowest possible.
Let's solve for n:
1600 = (2n - 1) v / (4L)
1600 = (2n - 1) / (4L)
2n - 1 = 1600 * (4L)
Since n cannot be a fractional value, we need to find the lowest value of n for which the equation holds true. Let's try n = 1:
2(1) - 1 = 1600 * (4L)
1 = 1600 * (4L)
1 = 6400L
L = 1 / 6400
So, the length L of the open pipe is 1 / 6400 meters. Now let's find the frequency of the lowest tone in a closed pipe of length 2L:
f = (2n - 1) v / (4 * (2L))
f = (2 * 1 - 1) v / (8L)
f = (1) v / (8L)
f = v / (8L)
Therefore, the frequency of the lowest tone in a closed pipe of length 2L is v / (8L), or v / (8 * (1 / 6400)), which simplifies to 800 Hz.
So, the answer is option c) 800 Hz.
2) As sound becomes louder, the amplitude increases, so e) amplitude does increase.
However, the frequency (a), velocity (b), period (c), and wavelength (d) of the sound do not change with changes in volume. Therefore, the answer is a) frequency.
3) In an open pipe, the fundamental frequency (the lowest tone) is directly proportional to the length of the pipe. Therefore, if the lowest tone to resonate in an open pipe of length L is 300 Hz, then any frequency that is not a multiple of 300 Hz (including harmonics) will not resonate in the same pipe.
Let's check the options:
a) 800 Hz = 300 Hz * (2.67)
b) 600 Hz = 300 Hz * (2)
c) 400 Hz = 300 Hz * (1.33)
d) 900 Hz ≠ Multiple of 300 Hz
As we can see, option d) 900 Hz is not a multiple of 300 Hz, so it will not resonate in the same pipe.
Therefore, the answer is d) 900 Hz.
To solve these questions, we need to apply the concepts of resonant frequencies in open and closed pipes.
1) The fundamental frequency of an open pipe (also known as the first harmonic) is given by the formula f1 = v / (2L), where f1 is the frequency of the lowest tone, v is the speed of sound, and L is the length of the pipe.
Given that the lowest tone to resonate in an open pipe of length L is 1600 Hz, we can rewrite the formula as v = 2L * f1.
Now, for a closed pipe, the fundamental frequency (first harmonic) can be determined using the formula f1 = v / (4L). Given that the length of the closed pipe is 2L, the frequency of the lowest tone that will resonate in the closed pipe is f1 = v / (8L).
To find the frequency of the lowest tone that will resonate in a closed pipe of length 2L, we substitute f1 into the formula: f = v / (8L) = (2L * f1) / (8L) = f1 / 4.
Therefore, the frequency is 1600 Hz / 4 = 400 Hz.
So, the answer to question 1 is c) 200 Hz.
2) The amplitude of a sound increases as it becomes louder, so e) amplitude is not the correct answer.
The frequency (a), velocity (b), period (c), and wavelength (d) of a sound all remain constant. Therefore, the answer is e) amplitude.
3) The formula for the fundamental frequency of an open pipe is f1 = v / (2L).
Given that the lowest tone to resonate in an open pipe of length L is 300 Hz, we can solve for the speed of sound v by multiplying the frequency f1 by 2L: v = 2L * f1 = 2L * 300 Hz
Now, let's check which of the given frequencies will not resonate in the same pipe:
a) 800 Hz: Since 800 Hz is greater than 2L * 300 Hz, it will resonate in the same pipe.
b) 600 Hz: 600 Hz is equal to 2L * 300 Hz, so it will resonate in the same pipe.
c) 400 Hz: 400 Hz is lower than 2L * 300 Hz, so it will resonate in the same pipe.
d) 900 Hz: Since 900 Hz is greater than 2L * 300 Hz, it will not resonate in the same pipe.
Therefore, the correct answer is d) 900 Hz.