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