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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: v2 = 144 -9x2.

 Column-I Column-II
i.Motion is simple harmonic of periodp.2π3units
ii. Maximum displacement from the fixed point isq.12 units
iii.Maximum velocity of the particler.27 units
iv.Magnitude of acceleration at a distance (S) 3 units from the fixed point iss.4 units

Now, match the given columns and select the correct option from the codes given below.

a
i-p, ii-q, iii-r, iv - s
b
i-s, ii-r, iii-q, iv -p
c
i-p, ii-s, iii-q, iv -r
d
i-r, ii-p, iii-s, iv-q

detailed solution

Correct option is C

For (i): v2 = 144-9x2  ⇒ 2vdvdx=0-18x⇒a = -9x   ⇒ω2 =9 ⇒ω = 3Time period T = 2πω = 2π3 unitsFor (ii): ∵ v2≥0 ∴144-9x2≥0 ⇒x2≤16 ⇒x≤4⇒Amplitude = 4 unitsFor (iii): Maximum velocity = Aω = (4)(3) = 12 unitsFor (iv): At x = 3 units, a = -9x = -27 units

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