Question

Stationary waves of frequency 200 hz are formed in air. if the velocity of the wave is 360 m/s, the shortest distance between two anti-nodes is

A. 1.8 m
B. 3.6 m
C. 0.9 m
D. 0.45 m

Answer 1

Using the formula for wave speed (v = fλ), with a frequency of 200 Hz and wave velocity of 360 m/s, the wavelength is calculated as 1.8 m. The shortest distance between two antinodes, which is half the wavelength, is therefore 0.9 m. Therefore, the correct option is C.

Step-by-step explanation:

The student asks about the shortest distance between two antinodes in a stationary wave, given a frequency of 200 Hz and a wave velocity of 360 m/s. To solve this, we can use the formula for wave speed (v = fλ), where v is the velocity, f is the frequency, and λ is the wavelength. Since antinodes on a standing wave are separated by half a wavelength (λ/2), we need to find the wavelength first and then divide it by 2.

Using the given data: v = 360 m/s, f = 200 Hz, we get λ = v/f = 360 m/s / 200 Hz = 1.8 m. Thus, the distance between two antinodes, which is λ/2, is 1.8 m / 2 = 0.9 m.

The correct answer is therefore C. 0.9 m.