In the given progressive wave $\mathrm{y}=5\mathrm{sin}\left(100\mathrm{\pi t}-0.4\mathrm{\pi x}\right)$ where y and x are in metre, t is in seconds. What is:

1. amplitude?

2. wavelength?

3. frequency?

4. wave velocity?

5. particle velocity amplitude?

Hint: Use the standard equation of the wave.
Step 1: Compare the given equation with the standard equation of the wave.
The standard equation of a progressive wave is given by:
$\mathrm{y}=\mathrm{asin}\left(\mathrm{\omega t}-\mathrm{kx}+\mathrm{\varphi }\right)$
This is travelling along positive x-direction.
Given equation is, $\mathrm{y}=5\mathrm{sin}\left(100\mathrm{\pi t}-0.4\mathrm{\pi x}\right)$
Comparing with the standard equation;
1. Amplitude = 5m
2. $\mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}=0.4\mathrm{\pi }$

3.
$\mathrm{\omega }=100\mathrm{\pi }$

4. Wave velocity, $\mathrm{\nu }=\frac{\mathrm{\omega }}{\mathrm{k}},$where k is wave number and $\mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}$.
$\mathrm{v}=\frac{100\mathrm{\pi }}{0.4\mathrm{\pi }}=\frac{1000}{4}$
5.
$\mathrm{y}=5\mathrm{sin}\left(100\mathrm{\pi t}-0.4\mathrm{\pi x}\right)$

                                                     .....(i)
From Eq. (i):
$\frac{\mathrm{dy}}{\mathrm{dt}}=5\left(100\mathrm{\pi }\right)\mathrm{cos}\left[100\mathrm{\pi t}-0.4\mathrm{\pi x}\right]$

For particle, velocity amplitude=${\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}_{max}$
which will be for ${\left\{\mathrm{cos}\left[100\mathrm{\pi t}-0.4\mathrm{\pi x}\right]\right\}}_{max}=1$