Question

A 1.25 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 118 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 10.2 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

Answer 1

Mass of the block M = 1.25*10^-2 Kg

Spring constant K = 118 N/m

Initial speed u = 10.2 m/s

Let A be the amplitude of the block.

According to law of conservation of energy

                              (1/2)Mu^2 = (1/2)KA^2

                                          A = Sqrt(Mu^2/K)

                                             = 0.104 m

Answer 2

From the law of conservation of energy
(1/2)mv^2 = (1/2)kA^2

So the amplitude of the resulting simple harmonic motion is
A = (mv^2)/(k)
= (1.25 x 10^2 kg)(10.2 m/s)^2/(118 N/m)
= 110.2118 m

Answer 3

Given that mass of the block is m = 1.25 *10^-2 kg

spring constant is k = 118 N /m

initial speed of the block is v = 10.2 m/s  

               v = A √ k / m

             10.2 m/s = A √ 118 / 1.25*10^-2 kg

            A = 0.10 m