Question

The $eqn$ of SHPW Ps given by $y = 0.05 \sin \pi (20t - \frac{x}{6})m$ balculate the displacement of a particle at 5m from the origin & at the Instant 0.1sec. Also Find the phase different betn two partical seperated by 5m.

Answer 1

• Displacement at 5 m, t = 0.1 s: $-0.025\,\text{m}$
• Phase difference between two particles separated by 5 m: $\frac{5\pi}{6}\,\text{radians}$

Answer 2

Solution:

1. Particle Displacement Calculation

The wave equation is given by

$$y = 0.05 \sin \pi\left(20t - \frac{x}{6}\right)$$

Substitute $x = 5\,\text{m}$ and $t = 0.1\,\text{s}$:

$$\text{Argument} = \pi\left(20(0.1) - \frac{5}{6}\right) = \pi\left(2 - \frac{5}{6}\right) = \pi \left(\frac{12-5}{6}\right) = \frac{7\pi}{6}$$

Then,

$$y = 0.05 \sin\frac{7\pi}{6}$$

We know that $\sin\frac{7\pi}{6} = -\frac{1}{2}$. Thus,

$$y = 0.05 \times \left(-\frac{1}{2}\right) = -0.025\,\text{m}$$

2. Phase Difference Between Two Particles Separated by 5 m

The phase part in the wave is

$$\phi = \pi\left(20t - \frac{x}{6}\right)$$

The phase difference for two points separated by $\Delta x = 5\,\text{m}$ is given by the change in the spatial term:

$$\Delta \phi = \pi\left(\frac{(x+5)}{6} - \frac{x}{6}\right) = \pi\left(\frac{5}{6}\right) = \frac{5\pi}{6}\,\text{radians}$$