Question

A light spring has unstressed length of 16.5 cm. It is described by Hooke’s law with spring constant 5.30 N/m. The spring is placed horizontally on a table. One end of the horizontal spring is held on a fixed vertical axle, and the other end is attached to a puck of mass m that can move without friction over the horizontal surface. The puck is set into motion in a circle with a period of 1.80 s.

a. Find and expression for the extension of the spring x as it depends on m.

b. Evaluate x for m = 0.075 kg

c. Evaluate x for m = 0.150 kg

d. Evaluate x for m = 0.225 kg.

e. Describe the pattern of variation of x as it depends on m.

Answer 1

Consider the given data: Unstressed length of light spring, d=16.5 cm=0.165 m Spring constant, K= 5.30 N/m Time period, T = 1.80 s Assume, the d is the length of the unstressed light spring. Since, the spring is held constant a fixed vertical axle. Assume, the extension in the wire is (d+x) Consider the expression for the extension of the spring F=Kx Consider the expression for the Newton's second law of motion. F =ma Equate equation (1) and (2).

Equate equation (1) and (2). ma Kx Since, the perform circular motion. 2m Substitute the values in above equation. 2TT Kx Simplify the above expression by substituting the value of r

4×0.075×π×0.165 530-4x 0.075x ㎡ 0.1507 4.386 0.034m = 3.43cm Calculate the extension of the spring when the mass is 0.150 kg 4×0.150×㎡× 0.165 (18) 530-4x 0.150×㎡ 0.30157 3.4722 0.0868 m = 8.685cm

Calculate the extension of the spring when the mass is 0.225 kg 4× 0.225 × ㎡ × 0.165 5.30 -4× 0.225 × ㎡ 2 0.4523 2.5584 =0.176 m 17.67 cm