A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 8.50 N is applied. A 0.530-kg particle rests on a frictionless h

asked 2020-11-14
A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 8.50 N is applied. A 0.530-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (Assume that the direction of the initial displacement is positive.)
(a) What is the force constant of the spring? 280 N/m
(b) What are the angular frequency (?), the frequency, and the period of the motion?
? = 23.121 rad/s
f = 3.6817 Hz
T = 0.27161 s
(c) What is the total energy of the system? 0.35 J
(d) What is the amplitude of the motion? 5 cm
(e) What are the maximum velocity and the maximum acceleration of the particle?
(f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s.
(g) Determine the velocity and acceleration of the particle when t = 0.500 s. (Indicate the direction with the sign of your answer.)
v = _________________ \(\displaystyle\frac{{m}}{{s}}\)
a = _________________ \(\displaystyle\frac{{m}}{{s}^{{{2}}}}\)

Expert Answers (1)


e) \(\displaystyle{v}_{{\max}}={A}{w}={1.1561}\frac{{m}}{{s}}\)
f) \(\displaystyle{x}={A}{\cos{{\left({w}{t}\right)}}}={5}{\cos{{\left({23.121}\cdot{0.5}\right)}}}={2.677}{c}{m}\)
g) \(\displaystyle{v}=-{A}{w}{\sin{{\left({w}{t}\right)}}}={0.9765}\frac{{m}}{{s}}\)
a) F=kx.
b) \(\displaystyle{W}^{{{2}}}={\frac{{{k}}}{{{m}}}}\)
c) \(\displaystyle{W}={\frac{{{2}\pi}}{{{T}}}}\)
Frequency \(\displaystyle={\frac{{{1}}}{{{p}{e}{r}{i}{o}{d}}}}\)
d) \(\displaystyle{A}={5.00}{c}{m}.\)
e) \(\displaystyle{V}=-{A}{W}{\sin{{W}}}{t}\)
max velocity is when t=1.779406359 s
max acceleration is when t =0
f) \(\displaystyle{x}={A}{\cos{{W}}}{t}\)
x=0.9185806285 m
g) \(\displaystyle{v}=-{A}{w}{\sin{{\left({w}{t}\right)}}}={0.9765}\frac{{m}}{{s}}\)

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