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 List - I List - II (P) Displacement of the particle in 3 2 π sec (1) 2 π /3 units (Q) Motion is simple harmonic with time period (2) Less than distance moved by it (R) Maximum displacement from the fixed point is (3) 27 units (S) Magnitude of acceleration at a distance 3 cm from the fixed point is (4) 4 units

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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 : $v^{2} = 144 - 9 x^{2}$

List - I	List - II
(P) Displacement of the particle in $\frac{2 π}{3}$ sec	(1) $2 π /3$ units
(Q) Motion is simple harmonic with time period	(2) Less than distance moved by it
(R) Maximum displacement from the fixed point is	(3) 27 units
(S) Magnitude of acceleration at a distance 3 cm from the fixed point is	(4) 4 units

Options:

Answer:

Solution:

Compare $v^{2} = 144 - 9 x^{2}$ with the SHM form $v^{2} = ω^{2} (A^{2} - x^{2})$ .

This gives $ω = 3 rad/s$ and amplitude $A = 4 m$ .

Evaluating each List-I quantity with these values matches option (A).

Answer: A

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Kinematics

⏱ 2mℹ️ Source: QB

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