A particle executes simple harmonic motion according to the displacement equation `y = 10 cos(2 pi t + (pi)/(6))cm` where `t` is in second The velocity of the particle at `t = 1/6` second will be

Options

A. `-6.28ms^(-1)`
B. `-0.628ms^(-1)`
C. `0.628ms^(-1)`
D. `6.28ms^(-1)`

Explanation

To find the velocity of the particle executing simple harmonic motion described by the displacement equation \( y = 10 \cos(2 \pi t + \frac{\pi}{6}) \) cm at \( t = \frac{1}{6} \) seconds, we can follow these steps:

### Step 1: Differentiate the displacement equation
The velocity \( v \) of the particle is the derivative of the displacement \( y \) with respect to time \( t \). Thus, we need to compute:
\[ v(t) = \frac{dy}{dt} = \frac{d}{dt}[10 \cos(2 \pi t + \frac{\pi}{6})] \]

### Step 2: Apply the chain rule
Using the chain rule for differentiation, we have:
\[ v(t) = 10 \cdot (-\sin(2 \pi t + \frac{\pi}{6})) \cdot \frac{d}{dt}(2 \pi t + \frac{\pi}{6}) \]
Since the derivative of \( 2 \pi t + \frac{\pi}{6} \) with respect to \( t \) is \( 2 \pi \), we can simplify:
\[ v(t) = -10 \cdot \sin(2 \pi t + \frac{\pi}{6}) \cdot 2 \pi \]
\[ v(t) = -20 \pi \sin(2 \pi t + \frac{\pi}{6}) \]

### Step 3: Substitute \( t = \frac{1}{6} \) seconds
Now we substitute \( t = \frac{1}{6} \) into the velocity equation:
\[ v\left(\frac{1}{6}\right) = -20 \pi \sin\left(2 \pi \cdot \frac{1}{6} + \frac{\pi}{6}\right) \]
Calculating the argument of the sine function:
\[ 2 \pi \cdot \frac{1}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]
Thus, we have:
\[ v\left(\frac{1}{6}\right) = -20 \pi \sin\left(\frac{\pi}{3} + \frac{\pi}{6}\right) \]

### Step 4: Simplify the sine term
Now we simplify the sine term:
\[ \frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi}{6} + \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \]
Then:
\[ \sin\left(\frac{\pi}{2}\right) = 1 \]

### Step 5: Calculate the velocity
Substituting back, we find:
\[ v\left(\frac{1}{6}\right) = -20 \pi \cdot 1 = -20 \pi \]
Using \( \pi \approx 3.14 \):
\[ v\left(\frac{1}{6}\right) \approx -20 \cdot 3.14 = -62.8 \text{ cm/s} \]
Converting to meters per second:
\[ -62.8 \text{ cm/s} = -0.628 \text{ m/s} \]

### Final Answer
Thus, the velocity of the particle at \( t = \frac{1}{6} \) seconds is approximately:
\[ \boxed{-0.628 \text{ m/s}} \]