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A particle moves in a straight line. If u is the velocity at a distance x from a fixed point on the line and $v^{2} = a - {bx}^{2},$ where a and b arc constants, then:
(i) the motion continues along the positive r-direction only
(ii) the motion is simple harmonic
(iii) the particle oscillates with a frequency equal to $\frac{\sqrt{b}}{2 π}$
(iv) the total energy of the particle is ma

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(i, ii)

(ii, iii)

(iii, iv)

(ii, iv)

answer is B.

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Detailed Solution

$v^{2} = a - {bx}^{2} \Rightarrow 2 v dv = 0 - b (2 x dx) \Rightarrow v \frac{dv}{dx} = - bx \Rightarrow \frac{d^{2} x}{{dx}^{2}} + bx = 0 \Rightarrow motion is SHM and ω^{2} = b \Rightarrow f = \frac{ω}{2 π} = \frac{\sqrt{2}}{2 π}$

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