Question

The speed of waves on a string is 97 m/s. If the frequency of standing waves is 485Hz, how far apart are the two adjacent nodes?

Two sig figs and proper units

Answer 1

Fundamental mode of vibration occurs when there is a node at the closed end and an anti node at the open end.

In closed organ pipe end is essentially a node and the open end is anti node.

As the distance between node and anti node is y/2, and distance between a node and anti node is y/4

The frequency of the fundamental mode is also called first harmonic

n₁ = v/y = v/4L= fundamental frequency (because y = 4L)

but v = (e/d)^1/2, where e is bulk modulus and d is density of the air, L is length of pipe

n₁ = 1/4L*(e/d)^1/2

Similarly for second mode of vibration, the frequency is called second harmonics

n₂ = v/y or

n₂ = 3/4L(e/d)^1/2 = 3n₁ (because L = y/2 + y/4)

Similarly for third mode

n₃ = 5/4L(e/d)^1/2 or (because L = y/2+y/2+y/4)

n₃ = 5n₁

Thus n₁:n₂:n₃:.......

1:3:5:.......

In case of open organ pipe

L = y/2 and v = n₁y

n₁ = v/2L = 1/2(e/d)^1/2

this is fundamental frequency or first harmonic. Similarly frequency of second mode

n₂ = 2(v/2L) = 2*(1/2L)*(e/d)^1/2 = 2n₁

and frequency of third mode

n₃ = 3*(v/2L)

= 3*(1/2L)*(e/d)^1/2 = 3n₁

Thus

n₁:n₂:n₃:.........= 1:2:3:........

This concludes that the possible tones of an open pipe have frequencies in the ratio of natural numbers. The note emitted by the air column in an open pipe will have the full series of overtones and hence it is very rich in harmonic overtones.That's why the quality of a note emitted by an open pipe is always better and sweeter than that of closed pipe.

So if one side of organ pipe is closed off the frequency drop by half.