A) \[\frac{n}{4}\]
B) 4n
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 \[{{m}_{1}}=m,\text{ }{{m}_{2}}=4m\text{ };\text{ }{{n}_{1}}=n\](given) |
\[\therefore \] \[\frac{n}{{{n}_{2}}}=\sqrt{\frac{4m}{m}}=2\] |
or \[{{n}_{2}}=\frac{n}{2}\] |
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