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Exercises · 14.16

Q.A steel rod 100 cm100\ \text{cm} long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz2.53\ \text{kHz}. What is the speed of sound in steel?

Rajasthan RbseTextbookSubjective· 3mImportance★★★★★est
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The rod is clamped at the centre, so the centre is a node and the free ends are antinodes. This makes the rod length equal to half the wavelength. Using v=fλv = f \lambda, the speed of sound in steel comes out to 5060 m/s5060\ \text{m/s}.

The key idea here is to connect the physical setup — a rod clamped at the middle — to the standing wave pattern it supports. When you clamp a rod at its centre, that point cannot move. In a longitudinal wave, a point that cannot move is a displacement node. The free ends, on the other hand, can vibrate freely, so they are displacement antinodes.

So the rod is supporting a standing wave with a node at the centre and antinodes at both ends. That means the length of the rod is exactly half a wavelength: from one antinode to the next antinode is λ/2\lambda/2, and the rod spans exactly that distance.

For a rod clamped at the centre, fundamental mode:

L=λ2L = \frac{\lambda}{2}

Now let’s work through the numbers.

  1. Identify the given data

    Length of rod: L=100 cm=1.00 mL = 100\ \text{cm} = 1.00\ \text{m}

    Fundamental frequency: f=2.53 kHz=2530 Hzf = 2.53\ \text{kHz} = 2530\ \text{Hz}

  2. Relate length to wavelength

    As argued, for the fundamental mode:

L=λ2L = \frac{\lambda}{2}

So

λ=2L=2×1.00 m=2.00 m\lambda = 2L = 2 \times 1.00\ \text{m} = 2.00\ \text{m}

  1. Apply the wave speed formula For any wave, v=fλv = f \lambda. v=2530 Hz×2.00 m=5060 m/sv = 2530\ \text{Hz} \times 2.00\ \text{m} = 5060\ \text{m/s} …

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