The speed v of a particle moving along a straight line when it is at a distance x from a fixed point on the line is given by $v^{2} = 144 - 9 x^{2}$ . Select the correct alternatives

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The motion of the particle is SHM with time period $T = \frac{2 π}{3} u n i t$

The maximum displacement of the particle from the fixed point is 8 unit

The magnitude of the acceleration at a distance 3 unit from the fixed point is 24 unit

The motion of the particle is periodic but not simple harmonic

answer is A.

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Detailed Solution

The equation resembles equation for velocity of SHM. $v = ω \sqrt{A^{2} - x^{2}}$

Comparing given equation $v^{2} = 144 - 9 x^{2}$ . with equation of SHM, $v^{2} = A^{2} ω^{2} - x^{2} ω^{2}$
we get, $ω = 3, A = 4$ .
$T = \frac{2 π}{ω} = \frac{2 π}{3}$ unit
Maximum displacement = A = 4 unit. Hence option (B) is wrong
$|a| = ω^{2} |x| = 9 (3) = 27$ unit. Hence option (C) is wrong
Differentiating $v^{2} = 144 - 9 x^{2}$ , we get $a α - x$ . This makes it simple harmonic motion. Hence option (D) is wrong.