Question

For the stationary wave y=4sin(pi x /15) cos (96pi t) the distance between node and next anti node is

Answer 1

the distance bw node and next antinode is always lambda/4 ie 90degrees phase difference....

Answer 2

Answer: The distance between node and next anti node is 7.5 m

Explanation:

Given that,

The wave equation is

$y = 4\sin(\dfrac{\pi x}{15})\cos(96\pi t)$

we know that,

The general equation is

$y = a\sin(\dfrac{2\pi x}{\lambda})\cos(\dfrac{2\pi vt}{\lambda})$

On comparing general equation with given equation

$\dfrac{2\pi}{\lambda}=\dfrac{\pi}{15}$

$\lambda = 30$

The distance between node and next anti node is

$d = \dfrac{\lambda}{4}= \dfrac{30}{4}$

$d = 7.5 m$

Hence, The distance between node and next anti node is 7.5 m.