If `y=A sin(omegat-kx),` then the value of `(d^2y)/(dt^2)/(d^2y)/(dx^2)`

Options

A. `(k^2)/(omega ^2)`
B. `-(k^2)/(omega ^2)`
C. `(omega ^2)/(k^2)`
D. `-(omega ^2)/(k^2)`

Answer

To solve the problem, we need to find the value of \(\frac{d^2y}{dt^2} \div \frac{d^2y}{dx^2}\) for the function \(y = A \sin(\omega t - kx)\).

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = A \sin(\omega t - kx) \]

2. **Differentiate \(y\) with respect to \(t\)**:
\[ \frac{dy}{dt} = A \cos(\omega t - kx) \cdot \frac{d}{dt}(\omega t - kx) = A \cos(\omega t - kx) \cdot \omega \]
Thus,
\[ \frac{dy}{dt} = A \omega \cos(\omega t - kx) \]

3. **Differentiate again with respect to \(t\) to find \(\frac{d^2y}{dt^2}\)**:
\[ \frac{d^2y}{dt^2} = \frac{d}{dt}(A \omega \cos(\omega t - kx)) = A \omega \cdot (-\sin(\omega t - kx)) \cdot \frac{d}{dt}(\omega t - kx) \]
\[ \frac{d^2y}{dt^2} = -A \omega^2 \sin(\omega t - kx) \]

4. **Differentiate \(y\) with respect to \(x\)**:
\[ \frac{dy}{dx} = A \cos(\omega t - kx) \cdot \frac{d}{dx}(\omega t - kx) = A \cos(\omega t - kx) \cdot (-k) \]
Thus,
\[ \frac{dy}{dx} = -A k \cos(\omega t - kx) \]

5. **Differentiate again with respect to \(x\) to find \(\frac{d^2y}{dx^2}\)**:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-A k \cos(\omega t - kx)) = -A k \cdot (-\sin(\omega t - kx)) \cdot \frac{d}{dx}(\omega t - kx) \]
\[ \frac{d^2y}{dx^2} = -A k \cdot (-\sin(\omega t - kx)) \cdot (-k) = -A k^2 \sin(\omega t - kx) \]

6. **Now, compute the ratio \(\frac{d^2y}{dt^2} \div \frac{d^2y}{dx^2}\)**:
\[ \frac{d^2y}{dt^2} \div \frac{d^2y}{dx^2} = \frac{-A \omega^2 \sin(\omega t - kx)}{-A k^2 \sin(\omega t - kx)} \]
The \(-A\) and \(\sin(\omega t - kx)\) terms cancel out:
\[ = \frac{\omega^2}{k^2} \]

### Final Result:
\[ \frac{d^2y}{dt^2} \div \frac{d^2y}{dx^2} = \frac{\omega^2}{k^2} \]