For a body in S.H.M the velocity is given by the relation v=144−16x2m/sec.The maximum acceleration is

v=144−16x2=9×16−16x2v=432−x2=wA2−x2w=4rad/sec;A=3mamax=ω2A=(4)23=48m/s2

For a body in S.H.M the velocity is given by the relation v=144−16x2m/sec.The maximum acceleration is

v=144−16x2=9×16−16x2v=432−x2=wA2−x2w=4rad/sec;A=3mamax=ω2A=(4)23=48m/s2

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For a body in S.H.M the velocity is given by the relation $v = \sqrt{144 - 16 x^{2}} m / \sec$ .The maximum acceleration is

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12 m/sec²

16 m/sec²

36 m/sec²

48 m/sec²

answer is D.

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$\begin{array}{l} v = \sqrt{144 - 16 x^{2}} = \sqrt{9 \times 16 - 16 x^{2}} \\ v = 4 \sqrt{3^{2} - x^{2}} = w \sqrt{A^{2} - x^{2}} \\ w = 4 rad / \sec; A = 3 m \\ a_{max} = ω^{2} A = (4)^{2} 3 = 48 m / s^{2} \end{array}$

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For a body in S.H.M the velocity is given by the relation v=144−16x2m/sec.The maximum acceleration is

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For a body in S.H.M the velocity is given by the relation $v = \sqrt{144 - 16 x^{2}} m / \sec$ .The maximum acceleration is