The displacement of a simple harmonic oscillator is given by x = 4 cos `(2pit+pi//4)m`. Then velocity of the oscillator at t = 2 s is

Options

A. `4pisqrt(2)m//s`
B. `(4pi)/(sqrt(2))m//s`
C. `(sqrt(2)pi)/(4)m//s`
D. `(4sqrt(2))/(pi)m//s`

Explanation

To find the velocity of a simple harmonic oscillator given the displacement function, we will follow these steps:

### Step 1: Write down the displacement equation
The displacement of the simple harmonic oscillator is given by:
\[ x(t) = 4 \cos(2\pi t + \frac{\pi}{4}) \, \text{m} \]

### Step 2: Differentiate the displacement equation to find the velocity
The velocity \( v(t) \) is the time derivative of the displacement \( x(t) \):
\[ v(t) = \frac{dx}{dt} \]

Using the chain rule, we differentiate:
\[ v(t) = \frac{d}{dt}[4 \cos(2\pi t + \frac{\pi}{4})] \]

The derivative of \( \cos \) is \( -\sin \), so:
\[ v(t) = -4 \sin(2\pi t + \frac{\pi}{4}) \cdot \frac{d}{dt}(2\pi t + \frac{\pi}{4}) \]

The derivative of \( 2\pi t + \frac{\pi}{4} \) with respect to \( t \) is \( 2\pi \). Therefore:
\[ v(t) = -4 \cdot 2\pi \sin(2\pi t + \frac{\pi}{4}) \]
\[ v(t) = -8\pi \sin(2\pi t + \frac{\pi}{4}) \]

### Step 3: Substitute \( t = 2 \, \text{s} \) into the velocity equation
Now we need to find the velocity at \( t = 2 \, \text{s} \):
\[ v(2) = -8\pi \sin(2\pi(2) + \frac{\pi}{4}) \]
\[ v(2) = -8\pi \sin(4\pi + \frac{\pi}{4}) \]

### Step 4: Simplify the sine function
Since \( \sin(4\pi + \theta) = \sin(\theta) \), we can simplify:
\[ v(2) = -8\pi \sin(\frac{\pi}{4}) \]

### Step 5: Calculate \( \sin(\frac{\pi}{4}) \)
We know that:
\[ \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \]

### Step 6: Substitute back to find the velocity
Substituting this value back into the velocity equation:
\[ v(2) = -8\pi \cdot \frac{1}{\sqrt{2}} \]
\[ v(2) = -\frac{8\pi}{\sqrt{2}} \]

### Step 7: Rationalize the expression
To express this in a more standard form, we can rationalize the denominator:
\[ v(2) = -\frac{8\pi}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = -\frac{8\pi \sqrt{2}}{2} = -4\pi \sqrt{2} \]

### Step 8: State the final answer
The magnitude of the velocity (since we are often interested in the speed) is:
\[ |v(2)| = 4\pi \sqrt{2} \, \text{m/s} \]