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 :
\[n=\frac{1}{2\pi }\sqrt{\frac{g}{l}}\] \[n\propto \frac{1}{\sqrt{l}}\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{l}_{2}}}{{{l}_{1}}}=\sqrt{\frac{{{L}_{2}}}{2{{L}_{2}}}}}\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{1}{\sqrt{2}}\] \[\Rightarrow \] \[{{n}_{2}}=\sqrt{2}{{n}_{1}}\] \[{{n}_{2}}>{{n}_{1}}\] Energy \[E=2{{\pi }^{2}}m{{n}^{2{{a}^{2}}}}\] Energy is same \[\frac{a_{1}^{2}}{a_{2}^{2}}=\frac{{{m}_{2}}n_{2}^{2}}{{{m}_{1}}n_{1}^{2}}\] \[{{n}_{2}}>{{n}_{1}}\,and\,{{m}_{1}}={{m}_{2}}\] \[\therefore \] \[{{a}_{1}}>{{a}_{2}}\]
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