Question

A 4.90 kg object on a horizontal frictionless surface is attached to a spring with spring constant 1030 N/m. The object is displaced from equilibrium 50.0 cm horizontally and given an initial velocity of 10.1 m/s back toward the equilibrium position. (a) What is the frequency of the motion? Hz (b) What is the initial potential energy of the block–spring system? J (c) What is the initial kinetic energy? J (d) What is the amplitude of the motion?

Answer 1

To solve this problem we will apply the descriptive equations of the movement of a mass-spring system, which describe the angular velocity as a function of the spring constant and the mass, the kinetic energy as a function of its linear velocity and the potential energy as a function of the spring constant and spring compression.

Finally we will calculate the amplitude of the system through the relation of energy in the harmonic movement, let's start:

PART A) The angular velocity is described as

$\omega = \sqrt{\frac{k}{m}}$

here,

k = Spring constant

m = Mass

We have all of this values then the angular velocity is

$\omega = \sqrt{\frac{1030}{4.9}}$

$\omega = 14.49 \approx 14.5 rad/s$

Therefore the frequency of the motion would be given by

$\omega = 2\pi f$

$f = \frac{\omega}{2\pi}$

$f = \frac{14.5}{2\pi}$

$f = 2.3Hz$

PART B) Now the potential energy is given as function of the spring constant and the displacement squared, then

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

$U= \frac{1}{2} (1030)(50*10^{-2})^2$

$U = 128.75J$

PART C) Kinetic energy is given as function of the mass and velocity squared, then

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (4.9)(10.1)^2$

$KE = 249.92J$

PART D) The total energy of the system is equivalent to half the product between the amplitude squared and the spring constant, said total energy per conservation must be equivalent to the sum of the kinetic and potential energy previously found, therefore,

$E = \frac{1}{2}kA^2$

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

$128.75+249.92= \frac{1}{2}(1030)A^2$

Solving for A,

$A = 0.85m$

Therefore the amplitude of the motion is 0.85m