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The magnitude of displacement of a parti...

The magnitude of displacement of a particle moving in a circle of radius a with constant angular speed `omega` varries with time t is

A

`2asin omegat`

B

`2asin ((omegat)/2)`

C

`2acos omegat`

D

`2acos ((omegat)/2)`

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The correct Answer is:
To find the magnitude of displacement of a particle moving in a circle of radius \( a \) with constant angular speed \( \omega \), we can follow these steps: ### Step 1: Understand the motion The particle moves in a circular path with a constant angular speed. The angular displacement \( \theta \) at any time \( t \) can be expressed as: \[ \theta = \omega t \] ### Step 2: Identify the positions At time \( t = 0 \), let the particle be at point \( P \) on the circle. After time \( t \), the particle moves to point \( Q \). The displacement \( PQ \) is the straight line distance between these two points. ### Step 3: Use trigonometric relationships In the circular motion, we can use the coordinates of points \( P \) and \( Q \): - The coordinates of point \( P \) (initial position) are: \[ P = (a \cos(0), a \sin(0)) = (a, 0) \] - The coordinates of point \( Q \) (final position) after time \( t \) are: \[ Q = (a \cos(\theta), a \sin(\theta)) = (a \cos(\omega t), a \sin(\omega t)) \] ### Step 4: Calculate the displacement The displacement \( PQ \) can be calculated using the distance formula: \[ PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2} \] Substituting the coordinates: \[ PQ = \sqrt{(a \cos(\omega t) - a)^2 + (a \sin(\omega t) - 0)^2} \] \[ = \sqrt{(a (\cos(\omega t) - 1))^2 + (a \sin(\omega t))^2} \] \[ = a \sqrt{(\cos(\omega t) - 1)^2 + \sin^2(\omega t)} \] ### Step 5: Simplify the expression Using the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \): \[ (\cos(\omega t) - 1)^2 + \sin^2(\omega t) = (\cos^2(\omega t) - 2\cos(\omega t) + 1) + \sin^2(\omega t) \] \[ = 1 - 2\cos(\omega t) + 1 = 2 - 2\cos(\omega t) = 2(1 - \cos(\omega t)) \] Thus, we have: \[ PQ = a \sqrt{2(1 - \cos(\omega t))} \] ### Step 6: Use the double angle identity Using the identity \( 1 - \cos(x) = 2\sin^2\left(\frac{x}{2}\right) \): \[ PQ = a \sqrt{2 \cdot 2\sin^2\left(\frac{\omega t}{2}\right)} = a \cdot 2\sin\left(\frac{\omega t}{2}\right) \] Finally, we can express the magnitude of displacement as: \[ PQ = 2a \sin\left(\frac{\omega t}{2}\right) \] ### Final Answer The magnitude of displacement of the particle moving in a circle of radius \( a \) with constant angular speed \( \omega \) varies with time \( t \) as: \[ PQ = 2a \sin\left(\frac{\omega t}{2}\right) \]

To find the magnitude of displacement of a particle moving in a circle of radius \( a \) with constant angular speed \( \omega \), we can follow these steps: ### Step 1: Understand the motion The particle moves in a circular path with a constant angular speed. The angular displacement \( \theta \) at any time \( t \) can be expressed as: \[ \theta = \omega t \] ...
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