A) \[\frac{{{m}_{1}}g}{K}\]
B) \[\frac{{{m}_{2}}g}{K}\]
C) \[\frac{({{m}_{1}}+{{m}_{2}})g}{K}\]
D) \[\frac{({{m}_{1}}-{{m}_{2}})g}{K}\]
Correct Answer: A
Solution :
With mass \[{{m}_{2}}\] alone, the extension of the spring l is given as \[{{m}_{2}}g=kl\] ...(i) With mass \[({{m}_{1}}+{{m}_{2}})\], the extension \[{l}'\] is given by \[({{m}_{1}}+{{m}_{2}})g=k(l+\Delta l)\] ....(ii) The increase in extension is \[\Delta l\] which is the amplitude of vibration. Subtracting (i) from (ii), we get \[{{m}_{1}}g=k\Delta l\] or \[\Delta l=\frac{{{m}_{1}}g}{k}\]
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