An object with mass 0.170 kg is acted on by an elastic restoring force with forc
ID: 1497419 • Letter: A
Question
An object with mass 0.170 kg is acted on by an elastic restoring force with force constant 10.4 N/m. The object is set into oscillation with an initial potential energy of 0.130 J and an initial kinetic energy of 6.10 Times 10^-2 J. What is the amplitude of oscillation? What is the potential energy when the displacement is one-half the amplitude? At what displacement are the kinetic and potential energies equal? What is the value of the phase angle phi if the initial velocity is positive and the initial displacement is negative?Explanation / Answer
Here,
mass of object , m = 0.17 Kg
force constant , k = 10.4 N/m
potential energy , Ui = 0.130 J
initial kinetic energy , KEi = 6.1 *10^-2 J
part A)
let the amplitude of oscillation is A
0.5 * k * A^2 = total initial energy
0.5 * 10.4 * A^2 = 0.13 + 6.1 *10^-2
A = 0.192 m
the amplitude of oscillation is 0.192 m
part B)
at half the amplitude
Potential energy= 0.5 * k * (A/2)^2
Potential energy = 0.5 * 10.4 * (0.192/2)^2
Potential energy = 0.048 J
part c)
for kinetic energy = potential energy
potential energy = 0.5 * total energy
0.5 * k * x^2 = 0.5 * (0.130 + 6.1 *10^-2)
0.5 * 10.4 * x^2 = 0.5 * (0.130 + 6.1 *10^-2)
solving for x
x = 0.136 m
the displacement when potential energy and kinetic energy is equal is 0.136 m
part D)
for the initial phase angle
phase angle = arccos(sqrt(0.061/(0.061 + 0.13)))
phase angle = 55.6 degree
the phase angle is 55.6 degree
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