Gujarati
13.Oscillations
easy

A mass m is suspended from a spring of length l and force constant $K$. The frequency of vibration of the mass is ${f_1}$. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is ${f_2}$. Which of the following relations between the frequencies is correct

A

${f_1} = \sqrt 2 {f_2}$

B

${f_1} = {f_2}$

C

${f_1} =2 {f_2}$

D

${f_2} = \sqrt 2 {f_1}$

Solution

(d) When spring is cut into two equal parts then spring constant of each part will be $2K$ and so using $n \propto \sqrt K $, new frequency will be $\sqrt 2 $ times i.e. ${f_2} = \sqrt 2 \,{f_1}$.

Standard 11
Physics

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