Show that the equation,`y=a sin(omegat-kx)` satisfies the wave equation`(del^(2) y)/(delt^(2))=v^(2) (del^(2)y)/(delx^(2))`. Find speed of wave and the direction in which it is travelling.

Explanation

To show that the equation \( y = a \sin(\omega t - kx) \) satisfies the wave equation

\[ \frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2} \]

we will follow these steps:

### Step 1: Calculate \(\frac{\partial y}{\partial t}\)

Given the function:

\[ y = a \sin(\omega t - kx) \]

We differentiate \(y\) with respect to \(t\):

\[ \frac{\partial y}{\partial t} = a \cos(\omega t - kx) \cdot \frac{\partial}{\partial t}(\omega t - kx) = a \cos(\omega t - kx) \cdot \omega \]

Thus,

\[ \frac{\partial y}{\partial t} = a\omega \cos(\omega t - kx) \]

### Step 2: Calculate \(\frac{\partial^2 y}{\partial t^2}\)

Now we differentiate \(\frac{\partial y}{\partial t}\) with respect to \(t\):

\[ \frac{\partial^2 y}{\partial t^2} = \frac{\partial}{\partial t}(a\omega \cos(\omega t - kx)) = -a\omega^2 \sin(\omega t - kx) \]

So, we have:

\[ \frac{\partial^2 y}{\partial t^2} = -a\omega^2 \sin(\omega t - kx) \]

### Step 3: Calculate \(\frac{\partial y}{\partial x}\)

Next, we differentiate \(y\) with respect to \(x\):

\[ \frac{\partial y}{\partial x} = a \cos(\omega t - kx) \cdot \frac{\partial}{\partial x}(\omega t - kx) = a \cos(\omega t - kx) \cdot (-k) \]

Thus,

\[ \frac{\partial y}{\partial x} = -ak \cos(\omega t - kx) \]

### Step 4: Calculate \(\frac{\partial^2 y}{\partial x^2}\)

Now we differentiate \(\frac{\partial y}{\partial x}\) with respect to \(x\):

\[ \frac{\partial^2 y}{\partial x^2} = \frac{\partial}{\partial x}(-ak \cos(\omega t - kx)) = ak^2 \sin(\omega t - kx) \]

So, we have:

\[ \frac{\partial^2 y}{\partial x^2} = ak^2 \sin(\omega t - kx) \]

### Step 5: Substitute into the wave equation

Now we substitute \(\frac{\partial^2 y}{\partial t^2}\) and \(\frac{\partial^2 y}{\partial x^2}\) into the wave equation:

\[ -a\omega^2 \sin(\omega t - kx) = v^2 (ak^2 \sin(\omega t - kx)) \]

### Step 6: Simplify the equation

Dividing both sides by \(a \sin(\omega t - kx)\) (assuming \(a \neq 0\) and \(\sin(\omega t - kx) \neq 0\)) gives:

\[ -\omega^2 = v^2 k^2 \]

### Step 7: Relate \(v\) to \(\omega\) and \(k\)

From the above equation, we can express \(v\):

\[ v^2 = \frac{\omega^2}{k^2} \implies v = \frac{\omega}{k} \]

### Step 8: Determine the direction of wave propagation

The negative sign in the argument of the sine function \((\omega t - kx)\) indicates that the wave is traveling in the positive x-direction.

### Final Result

Thus, we have shown that the given equation satisfies the wave equation, and the speed of the wave is:

\[ v = \frac{\omega}{k} \]

The direction of wave propagation is in the positive x-direction.

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