Question

For the stationary wave $y=4sin\frac{\pi x}{15}cos96\pi t$ The distance between a node and the next antinode is

• A. $7.5cm$
• B. $15cm$
• C. $22.5cm$
• D. $30cm$

Answer 1

The correct option is A $7.5cm$

Given , $y=4sin\frac{\pi x}{15}cos96\pi t$
For nodes ,displacement is zero.
$\therefore sin\frac{\pi x}{15}=0$
$\Rightarrow \frac{\pi x}{15}=0,\pi ,2\pi$,
$\Rightarrow x=0,15,30$
For antinodes ,displacement is maximum.
$\therefore sin\frac{\pi x}{15}=1$
$\Rightarrow \frac{\pi x}{15}=\frac{\pi }{2},\frac{3\pi }{2},$
$x=\frac{15}{2},\frac{45}{2}$
Hence diference between noce and antinode
$=\frac{15}{2}-0=\frac{15}{2}=7.5cm$