1. Form of a travelling wave

For a wave travelling along the positive $x$-axis, the displacement is

$y(x,t)=A\sin(kx-\omega t)$

where:

• $A$ = amplitude
• $k$ = wave number
• $\omega$ = angular frequency

2. Find the amplitude

The question says each point moves to and fro through a total distance of $6\text{ cm}$.

In SHM, during one complete oscillation, a particle moves from one extreme to the other and back, covering total distance:

$4A$

So,

$4A=6\text{ cm}$

$A=\frac{6}{4}=1.5\text{ cm}=0.015\text{ m}$

But the options clearly use either $0.03$ m or $0.06$ m. In wave terminology for such questions, “to and fro through a total distance of 6 cm” is commonly interpreted as peak-to-peak distance $2A=6\text{ cm}$.

Thus,

$2A=6\text{ cm} \Rightarrow A=3\text{ cm}=0.03\text{ m}$

So we take

$A=0.03\text{ m}$

3. Find the angular frequency

Given frequency:

$f=245\text{ Hz}$

Angular frequency:

$\omega=2\pi f=2\pi(245)=490\pi$

Using $\pi\approx 3.14$,

$\omega\approx 490\times 3.14\approx 1538.6\ \text{rad s}^{-1}$

So,

$\omega\approx 1.5\times 10^3\ \text{rad s}^{-1}$

4. Find the wave number

Wave speed:

$v=300\text{ m s}^{-1}$

Using

$v=f\lambda$

$\lambda=\frac{v}{f}=\frac{300}{245}\approx 1.2245\text{ m}$

Now,

$k=\frac{2\pi}{\lambda}=\frac{2\pi f}{v}$

$k=\frac{2\pi(245)}{300}=\frac{490\pi}{300}\approx 5.13\ \text{rad m}^{-1}$

So,

$k\approx 5.1\ \text{rad m}^{-1}$

5. Write the wave equation

Since the wave travels along the positive $x$-axis,

$y(x,t)=A\sin(kx-\omega t)$

Substituting values:

$y(x,t)=0.03\sin(5.1x-(1.5\times 10^3)t)$

This matches Option B.

6. Check options

• A: wrong angular frequency
• B: correct amplitude, wave number, and angular frequency
• C: wrong amplitude
• D: wrong amplitude, wave number, and angular frequency

Therefore, the correct option is B.