question_answer

A weightless spring which has a force constant k oscillates with a frequency r% 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)  \[2n\]           

D)  \[n\]

Correct Answer: D

Solution :

The period of oscillation of a body of mass m suspended by a spring (force  constant k) is On cutting the spring in two   equal parts, the length of each part will remain half and the force constant will be doubled (2 k). Therefore, \[T'=2\pi \sqrt{\frac{2m}{2k}}=T\] Since, time period remains same in both cases, frequency of oscillation is also same.