The equation of a stationary wave along a stretched string is given by $y = 5 \sin \left( \frac{\pi x}{3} \right) \cos (40 \pi t)$. Here $x$ and $y$ are in $cm$ and $t$ is in seconds. The separation between two adjacent nodes is: (in $cm$)

  • A
    $1.5$
  • B
    $3$
  • C
    $6$
  • D
    $14$

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$A$ string vibrates according to the equation $y = 5 \sin \left( \frac{2\pi x}{3} \right) \cos (20\pi t)$,where $x$ and $y$ are in $cm$ and $t$ is in $sec$. The distance between two adjacent nodes is .... $cm$.

Give the values of the amplitude at a node and an anti-node in a stationary wave.

$A$ travelling wave represented by $y = A \sin (\omega t - kx)$ is superimposed on another wave represented by $y = A \sin (\omega t + kx)$. The resultant 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:

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At nodes in stationary waves,

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