Question

A 326 g object is attached to a spring and executes simple harmonic motion with a period of 0.250 s. If the total energy of the system is 5.83 J, find a) the maximum speed of the object, b) the force constant of the spring, and c) the amplitude of the motion.
*Provide full explanation and solution for each part.

Answer 1

Answer:

a) $v_{max}=5.98\frac{m}{s}$

b) $k=205.92\frac{N}{m}$

c) $A=0.24m$

Explanation:

a) In the equilibrium position of the system, that is when the spring is not elongated, the potential energy is zero. Therefore, the total energy of the system is the maximum kinetic energy:

$E=K_{max}\\E=\frac{mv_{max}^2}{2}\\v_{max}=\sqrt{\frac{2E}{m}}\\v_{max}=\sqrt{\frac{2(5.83J)}{0.326kg}}\\v_{max}=5.98\frac{m}{s}$

b) The force constant of the spring can be calculated from the natural frequency of the system:

$\omega^2=\frac{k}{m}\\k=m\omega^2$

Recall that $\omega=\frac{2\pi}{T}$, that is the distance traveled in one revolution divided into the time of one revolution. Replacing and solving for k:

$k=\frac{m4\pi^2}{T^2}\\k=\frac{(0.326kg)4\pi^2}{(0.25s)^2}\\k=205.92\frac{N}{m}$

c) The maximum speed is directly proportional to the amplitude of the motion:

$v_{max}=A\omega\\A=\frac{v_{max}}{\omega}\\A=\frac{(v_{max})T}{2\pi}\\A=\frac{(5.98\frac{m}{s})0.25s}{2\pi}\\A=0.24m$