Question

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 $v_2$. The value $\frac{v_2}{v_1}$ is:

• A. $2$
• B. $\sqrt{2}$
• C. $4$
• D. $\sqrt{3}$

Answer 1

The correct option is B $\sqrt{2}$

The frequency of oscillations of mass m and force constant k is, $v_1=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$
when the spring is cut to one-half of its length, its force constant is doubles $\left(2k\right)$.
Then frequency of oscillation of mass m will be

$v_2=\frac{1}{2\pi }\sqrt{\frac{2k}{m}}=\frac{v_2}{v_1}=\sqrt{2}$`