A) \[n/4\]
B) \[4n\]
C) \[n/2\]
D) \[2n\]
Correct Answer: C
Solution :
\[n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\Rightarrow n\propto \frac{1}{\sqrt{m}}\Rightarrow \frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{m}_{2}}}{{{m}_{1}}}}\] \[\Rightarrow \frac{n}{{{n}_{2}}}=\sqrt{\frac{4m}{m}}\Rightarrow {{n}_{2}}=\frac{n}{2}\]
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