A) \[{{x}^{2}}=a+bv\]
B) \[x=\sqrt{a+b{{v}^{2}}}\]
C) \[x=a-bv\]
D) \[x=\sqrt{a-b{{v}^{2}}}\]
Correct Answer: D
Solution :
Relation [d] which is\[x=\sqrt{a-b{{v}^{2}}}\]correctly represent the SHM because, velocity \[v=\omega \sqrt{{{a}^{2}}-{{x}^{2}}}\] Or \[x=\sqrt{{{a}^{2}}-{{v}^{2}}/{{\omega }^{2}}}\] Or \[x=\sqrt{{{a}^{2}}-b{{v}^{2}}}\] where \[b=\frac{1}{{{\omega }^{2}}}\]
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