Standing waves are produced in a $10 \, m$ long stretched string fixed at both ends. If the string vibrates in $5$ segments and the wave velocity is $20 \, m/s$,the frequency is ....... $Hz$.

For a stationary wave,$Y = 10 \sin \left( \frac{\pi x}{15} \right) \cos (48 \pi t) \text{ cm}$,the distance between a node and the successive antinode is (in $\text{ cm}$)

In a standing wave on a string rigidly fixed at both ends:

The following equations represent progressive transverse waves:
$Z_1 = A \cos(\omega t - kx)$,
$Z_2 = A \cos(\omega t + kx)$,
$Z_3 = A \cos(\omega t + ky)$,
$Z_4 = A \cos(2\omega t - 2ky)$.
$A$ stationary wave will be formed by superposing:

The transverse displacement of a string of a linear density $0.01 \text{ kg m}^{-1}$,clamped at its ends is given by $Y_{(x, t)} = 0.03 \sin \left(\frac{2 \pi x}{3}\right) \cos (60 \pi t)$,where $x$ and $y$ are in metres and time $t$ is in seconds. The tension in the string is: (in $\text{ N}$)

At a certain instant,a stationary transverse wave is found to have maximum kinetic energy. The appearance of the string at that instant is: