Question

The magnitude of displacement of a particle moving in a circle of radius 'a' with constant angular speed ω varies with time 't' as :

(A) 2 a sin ωt
(B) 2 a sin ωt/2
(C) 2 a cos ωt
(D) 2 a cos ωt/2

Please do specify the method by which u got ur answer...

Answer 1

The magnitude of displacement of a particle moving in a circle of radius 'a' with constant angular speed $\omega$ varies with time 't' as :

(A) 2 a sin $\omega t$;

(B) 2 a sin $\omega t / 2$;

(C) 2 a cos $\omega t$;

(D) 2 a cos $\omega t / 2$;

Please do specify the method by which u got ur answer...

Particle position:

$$\vec {R} = \operatorname {a c o s} (\omega t) \vec {\imath} + \operatorname {a s i n} (\omega t) \vec {j}$$

At $t = 0$:

$$\overrightarrow {R _ {0}} = a \vec {\imath}$$

Displacement:

$$\begin{array}{l} \vec {d} = \vec {R} - \overrightarrow {R _ {0}} = a \cos (\omega t) \vec {\imath} + a \sin (\omega t) \vec {j} - a \vec {\imath} = a (\cos (\omega t) - 1) \vec {\imath} + a \sin (\omega t) \vec {j} \\ d = \sqrt {\left(a (\cos (\omega t) - 1)\right) ^ {2} + (a \sin (\omega t)) ^ {2}} = a \sqrt {2 - 2 \cos (\omega t)} = a \sqrt {2 (1 - \cos (\omega t))} \\ 1 - \cos (\omega t) = 2 (\sin (\omega t / 2)) ^ {2} \\ d = a \sqrt {2 * 2 (\sin (\omega t / 2)) ^ {2}} = 2 a \sin (\omega t / 2) \\ \end{array}$$

Answer $d = 2a\sin (\omega t / 2)$ (B)

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