### 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 $=\frac{\pi}{15}$

## Show Answer

#### 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,

$v=\frac{\lambda}{T}=\frac{2 \pi / 3}{2 \pi / 15}=5$