Question

A block of mass m is suspended from a spring. Its frequency of oscillation is f. The spring is cut into two identical halves and the same block is suspended from one of the two pieces of the spring such that it just touches the other spring below in its equilibrium position. The frequency of small oscillations of the mass will be

• A. $f$
• B. $\sqrt{2}f$
• C. $2f$
• D. $4f(\sqrt{2}-1)$

Answer 1

The correct option is C

$2f$

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$, k = spring constant of the uncut spring.

When the spring is cut into two equal parts, the spring constant of each part = 2k and when they are arranged as shown, the equivalent spring constant becomes 4k, which when put in the formula above we get the frequency as $2f$.