• # question_answer                 A mass m is vertically suspended from a spring of negligible mass; the system oscillates with a frequency n. What will be the frequency of the system, if a mass 4 m is suspended from the same spring ? A)                 $\frac{n}{4}$   B)                 4 n          C)                 $\frac{n}{2}$   D)                                            2 n

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            ${{n}_{2}}=\frac{n}{2}$