$A$ stationary wave is represented by $y = A \sin(100t) \cos(0.01x)$,where $y$ and $A$ are in millimetres,$t$ is in seconds,and $x$ is in metres. The velocity of the constituent wave is ........... $m/s$.

$A$ string fixed at both its ends vibrates in $5$ loops as shown in the figure. The total number of nodes and antinodes are respectively:

For a stationary wave, $y = 12 \cos \left( \frac{\pi x}{10} \right) \sin(36\pi t) \text{ cm}$, the distance between a node and the successive antinode is (in $\text{ cm}$)

$A$ standing wave pattern is formed on a string. One of the waves is given by the equation $y_1 = a \cos(\omega t - kx + \pi/3)$. Find the equation of the other wave such that at $x = 0$,a node is formed.

The equation of a wave disturbance is given as: $y = 0.02 \cos \left( \frac{\pi}{2} + 50\pi t \right) \cos (10 x),$ where $x$ and $y$ are in meters and $t$ is in seconds. Choose the wrong statement:

$A$ standing wave exists in a string of length $150 \ cm$,which is fixed at both ends with rigid supports. The displacement amplitude of a point at a distance of $10 \ cm$ from one of the ends is $5\sqrt{3} \ mm$. The nearest distance between two points,within the same loop and having a displacement amplitude equal to $5\sqrt{3} \ mm$,is $10 \ cm$. Find the maximum displacement amplitude of the particles in the string (in $mm$).