Question

A 3.30 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 125 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 10.8 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

Answer 1

0.17547 m

Step-by-step explanation:

m = Mass of block =
`3.3* 10^(-2)\ kg`

v = Velocity of block = 10.8 m/s

k = Spring constant = 125 N/m

A = Amplitude

The kinetic energy of the system is conserved

`(1)/(2)mv^2=(1)/(2)kA^2\\\Rightarrow \\\Rightarrow A=\sqrt{(mv^2)/(k)}\\\Rightarrow A=\sqrt{(3.3* 10^(-2)* 10.8^2)/(125)}\\\Rightarrow A=0.17547\ m`

The amplitude of the resulting simple harmonic motion is 0.17547 m