• 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

    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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