question_answer

The frequency of oscillations of a mass m suspended by a spring is \[{{v}_{1}}\]. If the length of the spring is cut to one-half, the same mass oscillates with frequency \[{{v}_{2}}\]. Determine the value of \[{{v}_{2}}/{{v}_{1}}\].

Answer:

                Let k be the force constant of the full spring Then frequency of oscillation of mass m will be \[{{v}_{1}}=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\] When the spring is cut to one-half of its length, its force constant is doubled \[(2k)\]. Frequency of oscillation of mass m will be \[{{v}_{2}}=\frac{1}{2\pi }\sqrt{\frac{2k}{m}}\] \[\therefore \]  \[{{v}_{2}}/{{v}_{1}}=\sqrt{2}\]