Spacing between two successive nodes in a standing wave on a string is $x$. If the frequency of the standing wave is kept unchanged but the tension in the string is doubled,then the new spacing between successive nodes will become

$A$ wave of frequency $100 \, Hz$ is sent along a string towards a fixed end. When this wave travels back after reflection,a node is formed at a distance of $10 \, cm$ from the fixed end of the string. The speed of the incident (and reflected) wave is .... $m/s$.

The correct statement about stationary waves is that

$A$ standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by $y(x, t) = 0.5 \sin(\frac{5\pi}{4}x) \cos(200\pi t)$. What is the speed of the travelling wave moving in the positive $x$ direction in $m/s$? ($x$ and $t$ are in meter and second,respectively.)

The equation of a standing wave in a string fixed at both ends is given as $y = 2A \sin kx \cos \omega t$. The amplitude and frequency of a particle vibrating at the midpoint between an antinode and a node are respectively:

$A$ stationary wave is represented by $y = 12 \cos \left(\frac{\pi}{6} x\right) \sin (8 \pi t)$,where $x$ and $y$ are in $cm$ and $t$ is in seconds. The distance between two successive antinodes is (in $cm$)