Waves - Result Question 33
33. The transverse wave represented by the equation $y=4 \sin (\frac{\pi}{6}) \sin (3 x-15 t)$ has
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(a) amplitude $=4$
(b) wavelength $=4 \frac{\pi}{3}$
(c) speed of propagation $=5$
(d) period $=\frac{\pi}{15}$
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Answer:
Correct Answer: 33. (c)
Solution:
- (c) Given $y=4 \sin (\frac{\pi}{6}) \sin (3 x-15 t)$
Compare the given equation with standard form $y=A \sin [\frac{2 \pi x}{\lambda}-\frac{2 \pi t}{T}]$
where, $\frac{2 \pi}{\lambda}=3, \lambda=\frac{2 \pi}{3}$ and $\frac{2 \pi}{T}=15$
$T=\frac{2 \pi}{15}$
Speed of propagation,
$v=\frac{\lambda}{T}=\frac{2 \pi / 3}{2 \pi / 15}=5$
