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

• A. 3 cm
• B. 9 cm
• C. 6 cm
• D. 12 cm

Hint:

In spring–circular motion problems, remember that the centripetal force is provided by the spring force, and the radius always equals the natural length plus elongation.

Answer 1

The Correct Option is B

Step 1: Identify the force acting on the mass.
When a mass attached to a spring is whirled in a horizontal circle, the required centripetal force is provided by the spring force. Thus, \[ k x = m \omega^2 r \] where \( k \) is the spring constant, \( x \) is the elongation, \( \omega \) is the angular speed, and \( r \) is the radius of the circular motion.

Step 2: Write equations for the two given conditions.
Let the natural length of the spring be \( l \).
For angular speed \( \omega \), elongation \( x_1 = 1 \, \text{cm} \): \[ k(1) = m \omega^2 (l + 1) \] When the angular speed is doubled, \( \omega' = 2\omega \), elongation \( x_2 = 6 \, \text{cm} \): \[ k(6) = m (2\omega)^2 (l + 6) = 4 m \omega^2 (l + 6) \]

Step 3: Eliminate constants and solve for the natural length.
Dividing the second equation by the first: \[ \frac{6}{1} = \frac{4(l + 6)}{(l + 1)} \] \[ 6(l + 1) = 4(l + 6) \] \[ 6l + 6 = 4l + 24 \] \[ 2l = 18 \Rightarrow l = 9 \, \text{cm} \]

Step 4: Final conclusion.
The original (natural) length of the spring is \( 9 \, \text{cm} \).