A) Amplitude of \[B\] greater than \[A\]
B) Amplitude of \[B\] smaller than \[A\]
C) Amplitudes will be same
D) None of these
Correct Answer: B
Solution :
[b] \[n=\frac{1}{2\pi }\sqrt{\frac{g}{l}}\Rightarrow \propto \frac{1}{\sqrt{l}}\Rightarrow \frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{l}_{2}}}{{{l}_{1}}}}=\sqrt{\frac{{{L}_{2}}}{2{{L}_{2}}}}\] \[\Rightarrow \frac{{{n}_{1}}}{{{n}_{2}}}=\frac{1}{\sqrt{2}}\Rightarrow {{n}_{2}}=\sqrt{2}{{n}_{1}}\Rightarrow {{n}_{2}}>{{n}_{1}}\] Energy, \[E=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}=2{{\pi }^{2}}m{{n}^{2}}{{a}^{2}}\] \[\Rightarrow \frac{{{a}^{2}}_{1}}{{{a}_{2}}^{2}}=\frac{{{m}_{2}}{{n}_{2}}^{2}}{{{m}_{1}}{{n}_{1}}^{2}}\]\[(\therefore E\,is\,same)\] Given\[{{n}_{2}}>{{n}_{1}}\] and \[{{m}_{1}}={{m}_{2}}\Rightarrow {{a}_{1}}>{{a}_{2}}\]
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