A particle executes SHM according to equation `x=10(cm)cos[2pit+(pi)/(2)]`, where t is in seconds. The magnitude of the velocity of the particle at `t=(1)/(6)s` will be :-

Options

A. 24.7 cm/s
B. 20.5 cm/s
C. 28.3 cm/s
D. 31.4 cm/s

Answer

To solve the problem, we need to find the magnitude of the velocity of a particle executing simple harmonic motion (SHM) given by the equation:

\[ x = 10 \cos(2\pi t + \frac{\pi}{2}) \]

where \( t \) is in seconds.

### Step-by-Step Solution:

1. **Identify the displacement equation**:
The displacement \( x \) of the particle is given as:
\[ x = 10 \cos(2\pi t + \frac{\pi}{2}) \]

2. **Differentiate the displacement to find velocity**:
The velocity \( v \) is the time derivative of the displacement \( x \):
\[ v = \frac{dx}{dt} \]
Using the chain rule, we differentiate:
\[ v = -10 \cdot 2\pi \sin(2\pi t + \frac{\pi}{2}) \]
Simplifying this gives:
\[ v = -20\pi \sin(2\pi t + \frac{\pi}{2}) \]

3. **Use the sine addition formula**:
We can use the sine addition formula:
\[ \sin(a + b) = \sin a \cos b + \cos a \sin b \]
Here, \( a = 2\pi t \) and \( b = \frac{\pi}{2} \). Thus:
\[ \sin(2\pi t + \frac{\pi}{2}) = \sin(2\pi t)\cos(\frac{\pi}{2}) + \cos(2\pi t)\sin(\frac{\pi}{2}) = \cos(2\pi t) \]
Therefore, we have:
\[ v = -20\pi \cos(2\pi t) \]

4. **Substitute \( t = \frac{1}{6} \) seconds**:
Now, we need to find the velocity at \( t = \frac{1}{6} \):
\[ v = -20\pi \cos(2\pi \cdot \frac{1}{6}) = -20\pi \cos(\frac{\pi}{3}) \]

5. **Calculate \( \cos(\frac{\pi}{3}) \)**:
We know that:
\[ \cos(\frac{\pi}{3}) = \frac{1}{2} \]
Substituting this back into the equation gives:
\[ v = -20\pi \cdot \frac{1}{2} = -10\pi \]

6. **Calculate the numerical value**:
Using \( \pi \approx 3.14 \):
\[ v \approx -10 \cdot 3.14 = -31.4 \, \text{cm/s} \]

7. **Find the magnitude of the velocity**:
The magnitude of the velocity is:
\[ |v| = 31.4 \, \text{cm/s} \]

### Final Answer:
The magnitude of the velocity of the particle at \( t = \frac{1}{6} \) seconds is \( 31.4 \, \text{cm/s} \).