A) Amplitude of B is greater than chat of A
B) Amplitude of B is smaller than that of A
C) Amplitude will be same
D) None of the above
Correct Answer: B
Solution :
Frequency, \[n=\frac{1}{2\pi }\sqrt{\frac{g}{l}}\] \[\Rightarrow \]\[{{n}_{2}}=\sqrt{2}\,{{n}_{1}}\] \[\Rightarrow \]\[{{n}_{2}}>{{n}_{1}}\] 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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