The correct statement about stationary waves is that

Stationary waves are set up in an air column. The velocity of sound in air is $330 \ m/s$ and the frequency is $165 \ Hz$. The distance between the nodes is ... $m$.

$A$ $20 \ cm$ long string,having a mass of $1.0 \ g$,is fixed at both the ends. The tension in the string is $0.5 \ N$. The string is set into vibrations using an external vibrator of frequency $100 \ Hz$. Find the separation (in $cm$) between the successive nodes on the string.

$A$ string of length $1\,m$ and linear mass density $0.01\,kg/m$ is stretched to a tension of $100\,N$. When both ends of the string are fixed,the three lowest frequencies for standing waves are $f_1, f_2$,and $f_3$. When only one end of the string is fixed,the three lowest frequencies for standing waves are $n_1, n_2$,and $n_3$. Then:

$A$ wave represented by the equation $y = A \cos (kx - \omega t)$ is superimposed with another wave to form a stationary wave such that the point $x = 0$ is a node. The equation of the other wave is:

Find the wrong statement from the following about the equation of a stationary wave given by $Y = 0.04 \cos(\pi x) \sin(50 \pi t) \text{ m}$,where $t$ is in seconds.