In stationary waves,the distance between a node and its nearest antinode is $20 \ cm$. The phase difference between two particles having a separation of $60 \ cm$ will be

Two progressive waves are travelling towards each other with velocity $50 \,m/s$ and frequency $200 \,Hz$. The distance between two consecutive antinodes is (in $\,m$)

$A$ wave represented by the equation $y = a \cos (kx - \omega t)$ is superposed with another wave to form a stationary wave such that the point $x = 0$ is a node. The equation for the other wave is:

$A$ string fixed at both ends vibrates in a resonant mode with a separation of $2.0 \, cm$ between consecutive nodes. For the next higher resonant frequency,this separation is reduced to $1.6 \, cm$. The length of the string is .... $cm$.

In a stationary wave,all particles

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: