question_answer

A weightless spring which has a force constant k oscillates with a frequency n when a mass m is suspended from it. The spring is cut into two equal halves and a mass 2m is attached to one of the halves. The frequency of oscillation will not become:

A)  \[\sqrt{2n}\]                                    

B) \[\frac{n}{\sqrt{2}}\]

C) \[\frac{2\pi }{\sqrt{b}}\]                                              

D)  n

Correct Answer: D

Solution :

                The period of oscillation of a body of mass m suspended by a spring (force constant k) is                                   \[T=2\pi \sqrt{\frac{m}{k}}\] On cutting the spring in two equal parts, the length of each part will remain half and the force constant will be doubled (2k). Therefore,                 \[T=2\pi \sqrt{\frac{2m}{2k}}=T\] Since, time period remains same in both cases, frequency of oscillation is also same.