Question

Rank the rods shown in the figure in terms of their fundamental resonant frequency, from lowest frequency to highest frequency.

Rank the rods shown in the figure in terms of their fundamental resonant frequency, from lowest frequency to highest frequency.

Answer 1

Concepts and reason

The concepts used to rank the fundamental frequencies of the rods is the principle of standing wave in rods and the wave equation. By determining the length required for the least frequency of the wave, the wavelength corresponding to the fundamental frequency can be found. Using the wave equation, the fundamental frequency can be determined.

Fundamentals

When a longitudinal wave passing through a metal rod is reflected at the other end of the rod along the straight line, a stationary wave is set up in the rod. The fundamental frequency with which the rod resonates is the smallest frequency of resonance.

A stationary wave or a standing wave has points which are always at rest, called nodes and points that vibrate with maximum amplitude called antinodes.

The fundamental frequency of resonance in a metal rod clamped at one end results when a node is formed at the fixed end and an antinode is formed at the free end.

The distance between a node and an antinode is $\frac{\lambda }{4}$ , where, $\lambda$ is the wavelength of the wave.

If the wave travels with a speed v in the metal rod, then the wave equation can be written as,

$v = f\lambda$ ……(1)

Here, f is the frequency of the wave.

Fundamental frequency of resonance occurs in a metal rod when a node is formed at its fixed end and an antinode is formed at the free end. This is shown in the diagram below.

Figure 1

Since the distance between a node and an antinode is $\frac{\lambda }{4}$ , it can be seen from figure 1, that,

$L = \frac{\lambda }{4}$ ……(2)

Here, L is the length of the rod.

Rewrite the wave equation, $v = f\lambda$ , for f.

$f = \frac{v}{\lambda }$ ……(3)

From equations (2) and (3), determine the expression for the fundamental frequency.

$\begin{array}{c}\\f = \frac{v}{\lambda }\\\\ = \frac{v}{{4L}}\\\end{array}$

Therefore, it can be seen that,

$f \propto \frac{1}{L}$ ……(4)

From the given diagram, it can be seen that the rod A is the longest, followed by D, then by B and the shortest rod is C.

By equation (4) $f \propto \frac{1}{L}$ , it can be seen that the longer the rod is lesser is its fundamental frequency.

Therefore, the fundamental frequency of A is the least, followed by D, then B and the rod C with the shortest length has the greatest fundamental frequency.

Ans:

The ranking of the fundamental frequency of the rods from the least to the greatest is as follows: A,D,B,C