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A particle executes SHM according to equ...

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 :-

A

24.7 cm/s

B

20.5 cm/s

C

28.3 cm/s

D

31.4 cm/s

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To find the magnitude of the velocity of the particle executing Simple Harmonic Motion (SHM) given by the equation \( x = 10 \, \text{cm} \cos(2\pi t + \frac{\pi}{2}) \) at \( t = \frac{1}{6} \, \text{s} \), we will follow these steps: ### Step 1: Differentiate the position function to find the velocity function The velocity \( v \) of the particle is given by the derivative of the position \( x \) with respect to time \( t \): \[ v = \frac{dx}{dt} \] Given \( x = 10 \cos(2\pi t + \frac{\pi}{2}) \), we differentiate: \[ v = \frac{d}{dt}[10 \cos(2\pi t + \frac{\pi}{2})] \] ### Step 2: Apply the chain rule Using the chain rule, we differentiate: \[ v = 10 \cdot (-\sin(2\pi t + \frac{\pi}{2})) \cdot \frac{d}{dt}(2\pi t + \frac{\pi}{2}) \] The derivative of \( (2\pi t + \frac{\pi}{2}) \) with respect to \( t \) is \( 2\pi \): \[ v = 10 \cdot (-\sin(2\pi t + \frac{\pi}{2})) \cdot 2\pi \] \[ v = -20\pi \sin(2\pi t + \frac{\pi}{2}) \] ### Step 3: Simplify the sine term Using the sine addition formula, we know: \[ \sin(a + b) = \sin a \cos b + \cos a \sin b \] Thus, \[ \sin(2\pi t + \frac{\pi}{2}) = \sin(2\pi t)\cos(\frac{\pi}{2}) + \cos(2\pi t)\sin(\frac{\pi}{2}) = 0 + \cos(2\pi t) = \cos(2\pi t) \] So, \[ v = -20\pi \cos(2\pi t) \] ### Step 4: Substitute \( t = \frac{1}{6} \, \text{s} \) Now, we substitute \( t = \frac{1}{6} \): \[ v = -20\pi \cos(2\pi \cdot \frac{1}{6}) = -20\pi \cos(\frac{\pi}{3}) \] Since \( \cos(\frac{\pi}{3}) = \frac{1}{2} \): \[ v = -20\pi \cdot \frac{1}{2} = -10\pi \] ### Step 5: Calculate the magnitude of the velocity The magnitude of the velocity is: \[ |v| = 10\pi \, \text{cm/s} \] Using \( \pi \approx 3.14 \): \[ |v| \approx 10 \cdot 3.14 = 31.4 \, \text{cm/s} \] ### Final Answer The magnitude of the velocity of the particle at \( t = \frac{1}{6} \, \text{s} \) is approximately \( 31.4 \, \text{cm/s} \). ---
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