Published 11.05.2018 on subject Physics by Guest

A 2.25m-long pipe has one end open. among its possible standing-wave frequencies is 345hz; the next higher frequency is 483hz. find(a) the fundamental frequency and (b) the sound speed.

Answered by sound

We know that
f = nv / 4L
for closed tubes, where n is odd numbers (1, 3, 5, 7...).
The two given frequencies are both odd-number multiples of some fundamental frequency, and the second frequency has the next odd number after the first.

Testing a few odd numbers, we find that the n-numbers for 345Hz and 483Hz are 5 and 7, respectively.

345Hz / 5 = 69Hz
69Hz * 7 = 483Hz.

That 69Hz is the fundamental frequency.

Now that we have all variables except v from the equation above, we can solve for v:

v = 4*L*f / n
v = (4 * 2.25 * 345) / 5 = 621 m/s

or, using the second frequency and its n-value:

v = 4*L*f / n
v = (4*2.25*483) / 7 = 321 m/s.