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) $y=0.2 \sin [2 \pi(6 t+\frac{x}{60})]$
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Answer:
Correct Answer: 12. (a)
Solution:
- (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
$y=a \sin (\omega t-k x)$
$=a \sin (2 \pi n t-\frac{2 \pi x}{\lambda})$
$=a \sin 2 \pi(n t-\frac{x}{\lambda})=0.2 \sin 2 \pi(6 t-\frac{x}{60})$
