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

In a stationary wave,if the distance between a consecutive node and antinode is $5 \ cm$,then find the distance between two consecutive antinodes. (in $cm$)

$A$ wave travelling along the positive $x-$ axis is given by $y = A\sin (\omega t - kx)$. If it is reflected from a rigid boundary such that $80\%$ of the amplitude is reflected,then the equation of the reflected wave is:

Standing stationary waves can be obtained in an air column even if the interfering waves are

$A$ 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 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