`y=Asin omegat` where A and `omega` are constants. Find `(d^(2)y)/(dt^(2))`.

Options

A. `-Aomega^(2).cosomegat`
B. `-Aomega^(2).sinomegat`
C. `-Aomega^(3).sinomegat`
D. None of the Above

Explanation

To solve the problem of finding the second derivative of \( y = A \sin(\omega t) \), we will follow these steps:

### Step 1: Differentiate \( y \) with respect to \( t \)
Given:
\[ y = A \sin(\omega t) \]
We will differentiate \( y \) with respect to \( t \):
\[ \frac{dy}{dt} = A \frac{d}{dt}(\sin(\omega t)) \]
Using the chain rule, the derivative of \( \sin(\theta) \) is \( \cos(\theta) \), and we also need to multiply by the derivative of the inner function \( \omega t \):
\[ \frac{dy}{dt} = A \cos(\omega t) \cdot \frac{d}{dt}(\omega t) = A \cos(\omega t) \cdot \omega \]
Thus, we have:
\[ \frac{dy}{dt} = A \omega \cos(\omega t) \]

### Step 2: Differentiate \( \frac{dy}{dt} \) to find \( \frac{d^2y}{dt^2} \)
Now, we differentiate \( \frac{dy}{dt} \) with respect to \( t \):
\[ \frac{d^2y}{dt^2} = \frac{d}{dt}(A \omega \cos(\omega t)) \]
Again, using the chain rule, the derivative of \( \cos(\theta) \) is \( -\sin(\theta) \):
\[ \frac{d^2y}{dt^2} = A \omega \left(-\sin(\omega t)\right) \cdot \frac{d}{dt}(\omega t) = A \omega \left(-\sin(\omega t)\right) \cdot \omega \]
Thus, we have:
\[ \frac{d^2y}{dt^2} = -A \omega^2 \sin(\omega t) \]

### Final Result
The second derivative of \( y \) with respect to \( t \) is:
\[ \frac{d^2y}{dt^2} = -A \omega^2 \sin(\omega t) \]

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