Waves - Result Question 12

12. The equation for a transverse wave travelling along the positive $x$-axis with amplitude $0.2 m$, velocity $v=360 ms^{-1}$ and wavelength $\lambda=60 m$ can be written as

[2002]

(a) $y=0.2 \sin [2 \pi(6 t-\frac{x}{60})]$

(b) $y=0.2 \sin [\pi(6 t+\frac{x}{60})]$

(c) $y=0.2 \sin [\pi(6 t-\frac{x}{60})]$

(d) $y=0.2 \sin [2 \pi(6 t+\frac{x}{60})]$

Show Answer

Answer:

Correct Answer: 12. (a)

Solution:

  1. (a) As we know that, velocity $v=n \lambda$

$\Rightarrow$ frequency, $n=\frac{v}{\lambda}=\frac{360}{60}=6$

Amplitude, $a=0.2$

For a wave travelling along positive $x$-axis

$y=a \sin (\omega t-k x)$

$=a \sin (2 \pi n t-\frac{2 \pi x}{\lambda})$

$=a \sin 2 \pi(n t-\frac{x}{\lambda})=0.2 \sin 2 \pi(6 t-\frac{x}{60})$



NCERT Chapter Video Solution

Dual Pane