Question

For the stationary wave $Y=4sin\left(\begin{array}{c}\frac{\pi x}{15}\end{array}\right)cos\left(96\pi t\right)$, the distance between a node and the next anti-node is(in cm) :

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

Answer 1

The correct option is A $7.5$

The equation of stationary wave $Y=4sin(\frac{\pi x}{15})cos\left(96\pi t\right)$

On comparing with $Y=Asin\left(kx\right)cos\left(wt\right)$, we get $k=\frac{\pi }{15}$

$\therefore$ $\frac{2\pi }{\lambda }=\frac{\pi }{15}$ $⟹\lambda =30$

The distance between a node and the next anti-node is one-fourth of the wavelength of the wave.

$⟹$ $d=\frac{\lambda }{4}=\frac{30}{4}=7.5$ cm