The lowest tone to resonate in a pipe of length l that is closed at one end but open at the other end is 200 hz. which one of the following frequencies will not resonate in that pipe?

Respuesta :

The frequency of 400 Hz is not possible.

Given that:

Frequency = 200 Hz

Length = l

Let's suppose, the given frequencies are,

600 Hz, 1000 Hz, 1400 Hz, 1800 Hz, and 400 Hz respectively.

To determine the possible resonance frequencies, we need to calculate the fundamental frequency.

Now using the formula of fundamental frequency for pipe

Where, n = odd number

F = nv / 4L

To calculate the first overtone, use the above formula.

n = 3

F = nv/4L = 3 x v / 4L = 3 x 200 = 600 Hz

For the second overtone, Use the same formula.

n = 5,

F = nv / 4L = 5 x 200 = 1000 Hz

For the third overtone, use the formula

n = 7,

F = nv / 4L = 7 x 200 = 1400 Hz

For the fourth overtone, use the formula.

n = 9,

F = nv/ 4L = 9 x 200 = 1800Hz

Hence, The frequency of 400 Hz is not possible.

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