Question

A very small mass m is fixed to one end of a massless spring of constant k and normal length l. The spring and the mass are rotated about the other end of the spring with angular speed ω. Neglect the effect of gravity. Extension in the spring is

• A. Zero
• B. $\frac { \mathrm { ml } \omega ^ { 2 } } { \mathrm { k } + \mathrm { m } \omega ^ { 2 } }$
• C. mlω²
• D. $\frac { m \omega ^ { 2 } 1 } { k - m \omega ^ { 2 } }$

Answer 1

Answer: (D)

$\frac { m \omega ^ { 2 } 1 } { k - m \omega ^ { 2 } }$

Answer 2

Here spring force = centripetal force

∴ kx = m(l + x)ω² (where x is the extension in the length of the spring)

i.e., x = $\frac { \mathrm { ml } \omega ^ { 2 } } { \mathrm { k } - \mathrm { m } \omega ^ { 2 } }$