In the standing wave shown,particles at the positions $A$ and $B$ have a phase difference of

$A$ transverse displacement of a vibrating string is given by $y = 0.06 \sin \left( \frac{2 \pi}{3} x \right) \cos (120 \pi t)$. If the mass per unit length of the string is $4 \times 10^{-2} \ kg/m$,then the tension in the string will be: (in $N$)

$A$ string is vibrating in its fifth overtone between two rigid supports $2.4 \ m$ apart. The distance between successive node and antinode is (in $m$)

The vibration of a string fixed at both ends is described by $Y = 2 \sin(\pi x) \sin(100\pi t)$,where $Y$ is in $mm$,$x$ is in $cm$,and $t$ is in $sec$. Then:

$A$ string fixed at one end and free at the other is vibrating in its second overtone. The length of the string is $10 \ cm$ and the maximum amplitude of vibration of particles of the string is $2 \ mm$. Then the amplitude of the particle at $9 \ cm$ from the fixed end is

$A$ 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: