Question

A 16.6 kg object on a horizontal frictionless surface is attached to a spring with k = 2470 N/m. The object is displaced from equilibrium 28.1 cm horizontally and given an initial velocity of 12.6 m/s back toward the equilibrium position. What are (a) the motion's frequency, (b) the initial potential energy of the block-spring system, (c) the initial kinetic energy, and (d) the motion's amplitude?

Answer 1

To solve this problem it is necessary to apply the concepts related to Period in a spring, elastic potential energy and energy in simple harmonic movement.

From the definition we know that the period can be expressed as

$T = 2\pi \sqrt{\frac{m}{k}}$

Where

m= Mass

k = Spring constant

Our values are given as

m = 16.6Kg

K = 2470N/m

Therefore the period would be

$T = 2\pi \sqrt{\frac{m}{k}}$

$T = 2\pi \sqrt{\frac{16.6}{2470}}$

$T = 0.515s$

PART A) The frequency is the inverse of the period therefore

$f = \frac{1}{T}\\f = \frac{1}{0.515}\\f = 1.9417Hz$

PART B) Elastic potential energy depends on compression and elasticity therefore

$PE = \frac{1}{2}kx^2$

$PE = \frac{1}{2} (2470)(0.281)^2$

$PE = 97.51J$

PART C) Kinetic energy depends on mass and speed therefore

$KE = \frac{1}{2}mv^2\\KE = \frac{1}{2} (16.6)(12.6)^2\\KE = 1371.7J$

PART D) As energy is conserved, the total energy is equivalent to the dual ratio of the elasticity constant and the amplitude, mathematically,

$KE+PE = \frac{1}{2}kA^2$

$1371.7+97.51 = \frac{1}{2}2470A^2$

$A = \sqrt{\frac{1415*2}{2470}}$

$A = 1.07m$

Therefore the motion's amplitude is 1.07m