If `x=A cos 4t+B sin4t,` then `(d^(2)x)/(dt^(2))` is equal to

Options

A. `-16x`
B. `16x`
C. x
D. `-x`

Answer

To solve the problem, we need to find the second derivative of the function \( x = A \cos(4t) + B \sin(4t) \).

### Step-by-Step Solution:

1. **Identify the function**:
\[ x = A \cos(4t) + B \sin(4t) \]

2. **First Derivative**:
We differentiate \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = \frac{d}{dt}(A \cos(4t)) + \frac{d}{dt}(B \sin(4t)) \]
Using the chain rule:
\[ \frac{d}{dt}(\cos(kt)) = -k \sin(kt) \quad \text{and} \quad \frac{d}{dt}(\sin(kt)) = k \cos(kt) \]
We get:
\[ \frac{dx}{dt} = -4A \sin(4t) + 4B \cos(4t) \]

3. **Second Derivative**:
Now we differentiate \( \frac{dx}{dt} \) to find \( \frac{d^2x}{dt^2} \):
\[ \frac{d^2x}{dt^2} = \frac{d}{dt}(-4A \sin(4t) + 4B \cos(4t)) \]
Again using the chain rule:
\[ \frac{d}{dt}(\sin(kt)) = k \cos(kt) \quad \text{and} \quad \frac{d}{dt}(\cos(kt)) = -k \sin(kt) \]
We have:
\[ \frac{d^2x}{dt^2} = -4A(4 \cos(4t)) - 4B(4 \sin(4t)) \]
Simplifying this gives:
\[ \frac{d^2x}{dt^2} = -16A \cos(4t) - 16B \sin(4t) \]

4. **Factor out -16**:
We can factor out \(-16\):
\[ \frac{d^2x}{dt^2} = -16(A \cos(4t) + B \sin(4t)) \]
Since \( x = A \cos(4t) + B \sin(4t) \), we can substitute back:
\[ \frac{d^2x}{dt^2} = -16x \]

### Final Answer:
\[ \frac{d^2x}{dt^2} = -16x \]