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 \[{{v}_{2}}=\frac{1}{2\pi }\sqrt{\frac{2k}{m}}\] \[\therefore \] \[{{v}_{2}}/{{v}_{1}}=\sqrt{2}\]
You need to login to perform this action.
You will be redirected in
3 sec