Question

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/sf = 3.6817 HzT = 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?vmax = 1.1561 m/samax = 26.73 m/s2(f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s. ________________ cm(g) Determine the velocity and acceleration of the particle when t = 0.500 s. (Indicate the direction with the sign of your answer.)v = _________________ m/sa = ___________________ m/s2

Answer 1

Answer:

283.33 N/m

23.121 rad/s

3.67982 Hz

0.27175 second

5 cm

1.15605 m/s

26.72903205 m/s²

2.6767 cm

0.97643 m/s

-14.30946 m/s²

Explanation:

Spring constant is given by

$k=\dfrac{F}{x}\\\Rightarrow k=\dfrac{8.5}{0.03}\\\Rightarrow k=283.33\ N/m$

Spring constant is 283.33 N/m

Angular frequency is given by

$\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow \omega=\sqrt{\dfrac{283.33}{0.53}}\\\Rightarrow \omega=23.121\ rad/s$

Angular frequency is 23.121 rad/s

Frequency is given by

$f=\dfrac{\omega}{2\pi}\\\Rightarrow f=\dfrac{23.121}{2\pi}\\\Rightarrow f=3.67982\ Hz$

Frequency is 3.67982 Hz

Time period is given by

$T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{3.67982}\\\Rightarrow T=0.27175\ s$

Time period is 0.27175 seconds

Amplitude is x = 5 cm

Energy is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow E=\dfrac{1}{2}283.33\times 0.05^2\\\Rightarrow E=0.3541625\ J$

Maximum velocity is given by

$v_m=A\omega\\\Rightarrow v_m=0.05\times 23.121\\\Rightarrow v_m=1.15605\ m/s$

The maximum velocity is 1.15605 m/s

Maximum acceleration is given by

$a_m=A\omega^2\\\Rightarrow a_m=0.05\times 23.121^2\\\Rightarrow a_m=26.72903205\ m/s^2$

Maximum acceleration is 26.72903205 m/s²

Displacement is given by

$x=Acos(\omega t)\\\Rightarrow x=5cos(23.121\times 0.5)\\\Rightarrow x=2.6767\ cm$

Displacement is 2.6767 cm

Velocity is given

$v=-A\omega sin(\omega t)\\\Rightarrow v=-0.05\times 23.121sin(23.121\times 0.5)\\\Rightarrow v=0.97643\ m/s$

Velocity is 0.97643 m/s

Acceleration is given by

$a=-A\omega^2 cos(\omega t)\\\Rightarrow a=-0.05\times 23.121^2cos(23.121\times 0.5)\\\Rightarrow a=-14.30946\ m/s^2$

Acceleration is -14.30946 m/s²