A transverse harmonic wave on a string is described by
$\mathrm{y}\left(\mathrm{x},\mathrm{t}\right)=3.0\mathrm{sin}\left(36\mathrm{t}+0.018\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)$
where x and y are in cm and t is in sec. The positive direction of x is from left to right.

(a) The wave is travelling from right to left.

(b) The speed of the wave is 20 m/s.

(c) The frequency of the wave is 5.7 Hz.

(d) The least distance between two successive crests in the wave is 2.5 cm.

Choose the correct alternatives:

1. (a, b, d)

2. (a, b, c)

3. (b, c, d)

4. (a, c, d)

(2) Hint: Use the standard equation of wave.
Given, the equation is $\mathrm{y}\left(\mathrm{x},\mathrm{t}\right)=3.0\mathrm{sin}\left(36\mathrm{t}+0.018\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)$
Compare the equation with the standard form,
$\mathrm{y}=\mathrm{asin}\left(\mathrm{\omega t}+\mathrm{kx}+\mathrm{\varphi }\right)$
Step 1: Find the direction of the wave.
1. As the equation involves a positive sign with x, hence the wave is travelling from right to left. Hence, option 1 is correct.
Step 2: Find the velocity of the wave.
2. Given,
$⇒$                                  $\mathrm{\omega }=36⇒2\mathrm{\pi \nu }=36$
$\mathrm{\nu }=\mathrm{frequency}=\frac{36}{2\mathrm{\pi }}=\frac{18}{\mathrm{\pi }}$
$\mathrm{k}=0.018⇒\frac{2\mathrm{\pi }}{\mathrm{\lambda }}=0.018$

Step 3: Find the frequency of the wave.
3.
$\mathrm{\omega }=2\mathrm{\pi \nu }=36$
$⇒\mathrm{v}=\frac{36}{2\mathrm{\pi }}\mathrm{Hz}=\frac{18}{\mathrm{\pi }}=5.7\mathrm{Hz}$
Step 4: Find the wavelength of the wave.
4.
$\mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}=0.018$
$⇒\mathrm{\lambda }=\frac{2\mathrm{\pi }}{0.018}\mathrm{cm}$
Hence, the least distance between two successive crests