A) \[\frac{{{d}^{2}}x}{d{{t}^{2}}}+nx=0\]
B) \[\frac{{{d}^{2}}x}{d{{t}^{2}}}+{{n}^{2}}x=0\]
C) \[\frac{{{d}^{2}}x}{d{{t}^{2}}}-{{n}^{2}}x=0\]
D) \[\frac{{{d}^{2}}x}{d{{t}^{2}}}+\frac{1}{{{n}^{2}}}x=0\]
Correct Answer: B
Solution :
The displacement of x for all S.H.M. is given by \[x=a\cos (nt+b)\]Þ\[\frac{dx}{dt}=-na\sin (nt+b)\] Þ \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=-{{n}^{2}}a\cos (nt+b)\] Þ \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=-{{n}^{2}}x\] Þ \[\frac{{{d}^{2}}x}{d{{t}^{2}}}+{{n}^{2}}x=0\].
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