The position of a simple harmonic oscillator is given by `x(t) = (0.50) cos ((pi)/(3)t)`, where t is measured in seconds. What is the maximum velocity of this oscillator?

Options

A. 0.17m/s
B. 0.52m/s
C. 0.67m/s
D. 1.0m/s

Answer

To find the maximum velocity of the simple harmonic oscillator given by the equation \( x(t) = 0.50 \cos\left(\frac{\pi}{3} t\right) \), we will follow these steps:

### Step 1: Identify the position function
The position function of the oscillator is given as:
\[ x(t) = 0.50 \cos\left(\frac{\pi}{3} t\right) \]

### Step 2: Differentiate the position function to find velocity
The velocity \( v(t) \) is the derivative of the position function \( x(t) \) with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} \]
Using the chain rule, we differentiate:
\[ v(t) = \frac{d}{dt}\left[0.50 \cos\left(\frac{\pi}{3} t\right)\right] \]
The derivative of \( \cos \) is \( -\sin \), so:
\[ v(t) = 0.50 \cdot \left(-\sin\left(\frac{\pi}{3} t\right)\right) \cdot \frac{d}{dt}\left(\frac{\pi}{3} t\right) \]
Calculating the derivative of \( \frac{\pi}{3} t \):
\[ \frac{d}{dt}\left(\frac{\pi}{3} t\right) = \frac{\pi}{3} \]
Thus, we have:
\[ v(t) = -0.50 \cdot \frac{\pi}{3} \sin\left(\frac{\pi}{3} t\right) \]
This simplifies to:
\[ v(t) = -\frac{0.50\pi}{3} \sin\left(\frac{\pi}{3} t\right) \]

### Step 3: Find the maximum velocity
The maximum value of \( \sin\left(\frac{\pi}{3} t\right) \) is 1. Therefore, the maximum velocity \( v_{\text{max}} \) is:
\[ |v_{\text{max}}| = \frac{0.50\pi}{3} \cdot 1 = \frac{0.50\pi}{3} \]

### Step 4: Calculate the numerical value
Now, we calculate the numerical value of \( \frac{0.50\pi}{3} \):
\[ v_{\text{max}} = \frac{0.50 \cdot 3.14}{3} \approx \frac{1.57}{3} \approx 0.5233 \, \text{m/s} \]
Rounding to two decimal places, we get:
\[ v_{\text{max}} \approx 0.52 \, \text{m/s} \]

### Final Answer
The maximum velocity of the oscillator is approximately:
\[ \boxed{0.52 \, \text{m/s}} \]

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