A transverse harmonic wave on a string is described by

$y(x,t)=3.0\sin \left(36t+0.018x+\frac{\pi }{4}\right)$ where x and y are in cm and t is in sec. The positive direction of x is from left to right.

Given equation is given by $y(x,t)=3.0\sin \left(36t+0.018x+\frac{\pi }{4}\right)$

Compare the equation with the standard form.  

              $y = a\sin (\omega t + kx + \varphi )$

(i)  As the equation involves positive sign with x ,hence the wave is travelling from right to left. Hence, option (i) is correct.

(ii)  Given,  $\omega = 36\Rightarrow 2\mathrm{\pi }v = 36$

$\Rightarrow v=\mathrm{frequency}=\frac{36}{2\mathrm{\pi }}=\frac{18}{\mathrm{\pi }}$

                              $k=0.018 \Rightarrow \frac{2\pi }{\lambda }=0.018$

$\Rightarrow \frac{2\pi v}{v\lambda }=0.018\Rightarrow \frac{\omega }{v}=0.018 \left[\because 2\lambda \mathrm{v}=\omega \text{ and }v\lambda =v\right]$

$\Rightarrow \frac{36}{v}=0.018=\frac{18}{1000}$

$\Rightarrow v = 2000 \mathrm{cm}/\mathrm{s} = 20 \mathrm{m}/\mathrm{s}$

(iii)  $2\mathrm{\pi }v=36$

$\Rightarrow v=\frac{36}{2\pi }\mathrm{Hz}=\frac{18}{\pi }=5.7\mathrm{Hz}$

(iv)  $\frac{2\pi }{\lambda }=0.018 \Rightarrow \lambda =\frac{2\pi }{0.018}\mathrm{cm}$

                                       $$\begin{gathered} =\frac{2000\pi }{18}\mathrm{cm} \\ =\frac{20\pi }{18}\mathrm{m} \\ =3.48\mathrm{cm} \end{gathered}$$

Hence, least distance between two successive crests $= \lambda = 3.48 \mathrm{m}.$