Problem (1) (Optional for inclusion in the examination solution.) A simple harmo

Question

Problem (1) (Optional for inclusion in the examination solution.) A simple harmonic oscillator begins its motion from maximum positive displacement when a certain clock reads t0.000 s. The period of the simple harmonic oscillator is 2.38 s. When the given clock reads time t 0.204 s, the velocity of the oscillator is -0.314 m/s. (a) Please find the position of the oscillator when t 4.83 s (b) If m -2.75 kg, what is the magnitude of the force experienced by the mass when the mass is at the position of part (a)? Suppose that this object is in simple harmonic motion, because it is a 2.75 kg mass sliding on a frictionless horizontal table and attached to an ideal Hooke's Law massless horizontal spring. (c) What is the valueof the force constant of the spring?

Explanation / Answer

equation of motion x = A*cos(wt)

velocity v = dx/dt = -A*w*sin(wt )

acceleration a = dv/dt = -A*w^2*cos(wt) = -w^2*x

time period T = 2pi/w = 2.38 s

angular speed w(omega) = 2*pi/2.38 = 2.64 rad/s

at time t = 0.204

v = -0.314 m/s

-0.314 = -A*2.64*sin(2.64*0.204)

amplitude A = 0.232 m

(a)

at time t = 4.83 s
position of particle x = 0.232*cos(2.64*4.83) = 0.228 m

(b)

acceleration a = dv/dt = -w^2*x

force F = m*a = -m*w^2*x

magnitude of Force F = m*w^2*x = 2.75*2.64^2*0.228= 4.37 N

-----------------------

(c)

force constant K = m*w^2 = 19.2 N/m

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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