In a stationary wave,all particles are

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 ... $m$.

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

$A$ string is rigidly tied at two ends and its equation of vibration is given by $y = \sin(2\pi x) \cos(2\pi t)$. Then the minimum length of the string is .... $m$

The stationary wave $y = 2a \sin kx \cos \omega t$ in a stretched string is the result of the superposition of $y_1 = a \sin(kx - \omega t)$ and

$A$ 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 $mm$).