Question
The transverse wave represented by the equation $$y = 4\sin \frac{\pi }{6}\sin \left( {3\,x - 15\,t} \right)$$ has
A.
amplitude $$ = 4\,\pi $$
B.
wavelength $$ = \frac{{4\,\pi }}{3}$$
C.
speed of propagation = 5
D.
period $$ = \frac{\pi }{{15}}$$
Answer :
speed of propagation = 5
Solution :
The standard equation of transverse wave is
$$y = a\sin \left[ {\frac{{2\,\pi t}}{T} - \frac{{2\,\pi x}}{\lambda }} \right]\,......\left( {\text{i}} \right)$$
Given equation is
$$y = 4\sin \frac{\pi }{6}\sin \left( {15t - 3x} \right)\,......\left( {{\text{ii}}} \right)$$
Comparing Eq. (ii) with Eq. (i)
$$\eqalign{
& \frac{{2\,\pi }}{\lambda } = 3, \cr
& \therefore \lambda = \frac{{2\,\pi }}{\lambda }\,\,{\text{and}}\,\,\frac{{2\,\pi }}{T} = 15 \cr
& \therefore T = \frac{{2\,\pi }}{{15}} \cr} $$
$$\therefore $$ Speed of propagation wave, $$v = \frac{\lambda }{T} = \frac{{\frac{{2\pi }}{3}}}{{\frac{{2\pi }}{{15}}}} = 5$$