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.You need to login to perform this action.
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