A) \[\frac{n}{4}\]
B) 4 n
C) \[\frac{n}{2}\]
D) 2 n
Correct Answer: C
Solution :
Key Idea: Time period of oscillating system whether it is a simple pendulum of spring-mass system, is given by \[T=2\pi \sqrt{\left( \frac{displacement}{acceleration} \right)}\] Time period of spring-mass system is \[n=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{acceleration}{displacement}}\] \[n=\frac{1}{2\pi }\sqrt{\frac{g}{l}}\] ....(i) In case of vertical spring mass system, in equilibrium position \[kl=mg\Rightarrow \frac{g}{l}=\frac{k}{m}\] where \[l=\] extension in the spring and \[k=\] spring constant or force constant of spring \[\therefore \,\] From Eq. (i), we have \[n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\] or \[n\propto \,\,\frac{1}{\sqrt{m}}\] or \[\frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{m}_{2}}}{{{m}_{1}}}}\] but m1 = m, m2 = 4m ; n1 = n (given) \[\therefore \frac{n}{{{n}_{2}}}=\sqrt{\frac{4m}{m}}=2\] or \[{{n}_{2}}=\frac{n}{2}\]
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