Question

Stationary waves of frequency $200Hz$ are formed in the air. If the velocity of the wave is $360m/s$, the shortest distance between two antinodes is

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

Answer 1

The correct option is C

$0.9m$

Step 1: Given data

The shortest distance between two antinodes is equal to the half of wavelength.

Here $v=360m/s$ and $f=200Hz$

Step 2: Formula used

Let $\lambda =$wavelength

$\therefore \lambda =\frac{v}{f}$

Where $f$ is the frequency and $v$ is the velocity of the wave

Step 3: Calculate the shortest distance between the antinodes

$\therefore$$\lambda =\frac{v}{f}=\frac{360}{200}=1.8m$

So $\frac{\lambda }{2}=\frac{1.8}{2}=0.9m$

Hence, option C is correct.