Question

A mass 'm' is tied to one end of a spring and whirled in a horizontal circle with constant angular velocity. The elongation in the spring is 1 cm. If the angular speed is doubled, the elongation in the spring is 6 cm. The original length of the spring is ______.

Options

• 3 cm

• 12 cm

• 6 cm

• 9 cm

Answer 1

A mass 'm' is tied to one end of a spring and whirled in a horizontal circle with constant angular velocity. The elongation in the spring is 1 cm. If the angular speed is doubled, the elongation in the spring is 6 cm. The original length of the spring is 9 cm.

Explanation:

Let l be the original length of the spring.

Let the initial angular velocity be ω and the corresponding elongation e₁ = 1 cm.

When the angular velocity is doubled the elongation e₂ = 6 cm.

If k is the spring constant then we have

m(l + e₁)ω² = ke₁               ....(1)

and m(l + e₂)(2ω)² = ke₂

or m(l + e₂) × 4ω² = ke₂        ......(2)

Dividing Eq(1) by Eq(2), we get

`(l+e_1)/(4(l+e_2))=e_1/e_2`

solving we get l = 9 cm.