A mass m suspended by a spring oscillates with a frequency v₁. If the spring is cut to one-half of its original length then the same mass oscillates with frequency v₂. Find the ratio v₂/v₁ 

The frequency of oscillation of mass m and force constant k is ${\mathrm{\nu }}_1=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$  When the spring is cut to one-half of its length, its force constant is double (2k). Then frequency of oscillation of mass m will be

${\mathrm{\nu }}_2=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{2\mathrm{k}}{\mathrm{m}}}\therefore \frac{{\mathrm{\nu }}_2}{{\mathrm{\nu }}_1}=\sqrt{2}$