First slide

Standing waves

Question

If two anti-nodes are observed on either side of C, the frequency of the mode in which the rod is vibrating will be ( $y = 2 \times 10^{11} N / m^{2}, d = 8 \times 10^{3} kg / m^{3}$ ) length of the rod is 2.5 m

Moderate

A

1000 Hz
B

3000 Hz
C

7000 Hz
D

1500 Hz

Solution

Velocity of the longitudinal waves in the rod $= v = \sqrt{Y / d} = \sqrt{2 \times 10^{11} / 8000} = 5000 m / s$
The wavelength of the wave for the mode of vibration in which 2 anti-nodes occur, is $λ = \frac{1.25}{(3 / 4)} = \frac{5}{3} m$
Hence frequency of vibration $= n = \frac{V}{λ} = \frac{5000}{5 / 3} Hz = 3000 Hz$

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