Q2 - A massless spring having a spring constant k = 19 N/m hangs vertically. An

Question

Q2 - A massless spring having a spring constant k = 19 N/m hangs vertically. An object of mass m = 0.2 kg is attached to the free end of the spring and then released. Assume that the spring was unstretched before the body was released. Find (a) the maximum distance the object descends below the initial position, and (b) the frequency and amplitude of the resulting simple harmonic motion.

Q3 - An object attached to a spring experiences simple harmonic motion (define the potential energy of the object + spring system to be zero when it is in equilibrium). Question: (a) When the displacement of the object is one-half the maximum amplitude A, what fraction of the total energy is kinetic and what fraction is potential? (b) At what displacement x is the total energy half kinetic and half potential?

Explanation / Answer

Q2-F = mg = 0.2*9.8 = 1.96
F = kx = 1.96 = 19x
x = 0.103 m: this is the equilibrium position.
It will continue to go down this same distance again before it stops.
(a) 0.206 m
(b) amplitude = 0.103 m; frequency = (1/(2pi))sqrt(k/m) = (1/(2pi))sqrt(19/0.2) = 1.55 Hz

Q3-Total Energy: E=K+U=1/2kA^2
U=1/2kx^2
x=1/2A
U=1/2k(1/2A)^2-----replace displacement with 1/2A
U=1/2k(1/4A^2)
U=1/8kA^2
E/U=(1/2kA^2)/(1/8kA^2)=1/4--------Pot... energy is 1/4 of total Energy
K=3/4 of total energy

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.