Question

A metal rod of length 125 cm is clamped at its midpoint. If the speed of the sound in the metal is 5000ms\(^{-1}\), then the fundamental frequency of the longitudinal vibrations of the rod is

• A. 2 kHz
• B. 20 kHz
• C. 0.2 kHz
• D. 200 kHz

Hint:

For standing waves in rods: - Clamped end/point = Node. - Free end = Antinode. The simplest vibration mode (fundamental) will have the fewest possible nodes and antinodes. For a rod clamped at the center, the fundamental mode has a wavelength twice the length of the rod (\(\lambda = 2L\)).

Answer 1

The Correct Option is A

When a rod is clamped at its midpoint, the midpoint must be a node (a point of zero displacement).
The ends of the rod are free to vibrate, so they must be antinodes (points of maximum displacement).
For the fundamental frequency (the simplest mode of vibration), the pattern will be Antinode - Node - Antinode.
The distance between an antinode and the next node is always one-quarter of a wavelength (\(\lambda/4\)).
The total length of the rod, L, covers the distance from one antinode end to the central node, plus the distance from the central node to the other antinode end.
So, \( L = \frac{\lambda}{4} + \frac{\lambda}{4} = \frac{\lambda}{2} \).
This means the wavelength of the fundamental mode is \( \lambda = 2L \).
We are given the length of the rod, \( L = 125 \text{ cm} = 1.25 \text{ m} \).
So, the fundamental wavelength is \( \lambda = 2 \times 1.25 = 2.5 \) m.
The relationship between frequency (f), wavelength (\(\lambda\)), and the speed of the wave (v) is \( v = f\lambda \).
We can find the fundamental frequency as \( f = \frac{v}{\lambda} \).
We are given the speed of sound in the metal, \( v = 5000 \) m/s.
\( f = \frac{5000}{2.5} = \frac{50000}{25} = 2000 \) Hz.
The question asks for the answer in kHz.
\( 2000 \text{ Hz} = 2 \text{ kHz} \).