Q.For the travelling harmonic wave

[!FORMULA] $y(x, t) = 2.0 \cos 2\pi(10t - 0.0080\,x + 0.35)$

where $x$ and $y$ are in cm and $t$ in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of

The key idea is that the phase difference between two points in a travelling harmonic wave depends on the distance separating them and the wave number.

The given wave equation is $y(x, t) = 2.0 \cos 2\pi(10t - 0.0080\,x + 0.35)$.

This can be rewritten as $y(x, t) = 2.0 \cos(20\pi t - 0.016\pi x + 0.7\pi)$.

Comparing this with the standard form $y(x, t) = A \cos(\omega t - kx + \phi_0)$, we identify the angular wave number $k = 0.016\pi \ \text{cm}^{-1}$.

We also know that $k = \frac{2\pi}{\lambda}$, where $\lambda$ is the wavelength.

So, $\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.016\pi} = \frac{1}{0.008} = 125\ \text{cm} = 1.25\ \text{m}$.

Now, we calculate the phase difference for each case:

1. For $\Delta x = 4\ \text{m} = 400\ \text{cm}$: $\Delta\phi = (0.016\pi \ \text{cm}^{-1}) \times (400\ \text{cm}) = 6.4\pi \ \text{rad}$.
2. For $\Delta x = 0.5\ \text{m} = 50\ \text{cm}$: …

A travelling harmonic wave describes how a disturbance propagates through a medium. The "phase" of the wave at a particular point in space and time tells us about the stage of its oscillation. For a travelling wave, the phase changes as you move along the direction of propagation ($x$) and also as time ($t$) progresses.

When we talk about the phase difference between two points, we are usually interested in how much one point "leads" or "lags" the other in its oscillatory motion at the same instant in time. This spatial phase difference is directly proportional to the distance separating the two points. If two points are separated by one full wavelength ($\lambda$), they are in phase, meaning their phase difference is $2\pi$ radians. If they are separated by half a wavelength ($\lambda/2$), they are exactly out of phase, with a phase difference of $\pi$ radians.

The general form of a travelling harmonic wave is often written as:

where:

• $A$ is the amplitude.
• $\omega$ is the angular frequency ($2\pi f$, where $f$ is the frequency).
• $k$ is the wave number ($2\pi/\lambda$, where $\lambda$ is the wavelength).
• $\phi_0$ is the initial phase constant.

The argument of the cosine function, $(\omega t - kx + \phi_0)$, is the phase of the wave at position $x$ and time $t$.

For two points, $x_1$ and $x_2$, at the same time $t$, their phases are:

$\phi_1 = \omega t - kx_1 + \phi_0$

$\phi_2 = \omega t - kx_2 + \phi_0$

The phase difference, $\Delta\phi$, is then:

The magnitude of the phase difference is $|k \Delta x|$. Since we are typically interested in the magnitude of the phase difference, we use:

Let's apply this to the given problem.

1. Extract wave parameters from the given equation. The given wave equation is $y(x, t) = 2.0 \cos 2\pi(10t - 0.0080\,x + 0.35)$. To match it with the standard form $y(x, t) = A \cos(\omega t - kx + \phi_0)$, we first distribute the $2\pi$ inside the cosine argument:

By comparing this with the standard form, we can identify the wave number $k$:
$k = 0.016\pi$.
The units for $x$ are in cm, so the units for $k$ are $\text{rad/cm}$.

2. Calculate the wavelength ($\lambda$).

The wave number $k$ is related to the wavelength $\lambda$ by the formula $k = \frac{2\pi}{\lambda}$.

We can rearrange this to find $\lambda$:

Substitute the value of $k$:

It's often useful to work in meters for consistency, especially since some given distances are in meters.
$$\lambda = 125 \text{ cm} = 1.25 \text{ m}$$ …

Step 1: Expand $y=2.0\cos2\pi(10t-0.0080x+0.35)$ into $2.0\cos(20\pi t-0.016\pi x+0.7\pi)$ and read off the wave number $k=0.016\pi\text{ rad/cm}$.

Step 2: Wavelength $\lambda=2\pi/k=2\pi/(0.016\pi)=125\text{ cm}=1.25\text{ m}$.

Step 3: Phase difference between two points separated by $\Delta x$ on the same wave: $\Delta\phi=k\Delta x=(2\pi/\lambda)\Delta x$.

Step 4 (a): $\Delta x=4\text{ m}=400\text{ cm}$: $\Delta\phi=0.016\pi\times400=6.4\pi\text{ rad}$. …