A) Amplitude of B is smaller than A
B) Amplitude of B is greater than B
C) Amplitude will be same
D) None of the above
Correct Answer: A
Solution :
Frequency, \[n=\frac{1}{2\pi }\sqrt{\frac{g}{l}}\] or \[n\propto \frac{1}{\sqrt{l}}\] \[\therefore \] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{l}_{2}}}{{{l}_{1}}}}=\sqrt{\frac{{{l}_{2}}}{2{{l}_{2}}}}\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{1}{\sqrt{2}}\] \[\Rightarrow \] \[{{n}_{2}}=\sqrt{2{{n}_{1}}}\] \[\Rightarrow \] \[{{n}_{2}}>{{n}_{1}}\] Energy, \[E=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] and \[{{a}^{2}}\propto \frac{1}{m{{n}^{2}}}\] {\[\because \] E is same} \[\therefore \] \[\frac{a_{1}^{2}}{a_{2}^{2}}=\frac{{{m}_{2}}n_{2}^{2}}{{{m}_{1}}n_{1}^{2}}\] Given, \[{{n}_{2}}>{{n}_{1}}\] and \[{{m}_{1}}>{{m}_{2}}\] \[\Rightarrow \] \[{{a}_{1}}>{{a}_{2}}\]. So, amplitude of B is smaller than A,
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