A mass m = 0.017 kg is attached to a horizontal spring with constant k = 42.0 N/

Question

A mass m = 0.017 kg is attached to a horizontal spring with constant k = 42.0 N/m. Friction and the mass of the spring itself are negligible. At time t = 0, the mass is observed to be displaced 30.0 mm to the right of its equilibrium position, and to be moving to the right at 3.471 m/s. The motion is described by x = xmcos(?t + ?), with the +x direction to the right.

What is the value of xm?

What is the value of the phase constant ?? Enter an angle between -? rad and +? rad.

What is the total energy in the oscillation of the spring and mass system?

Explanation / Answer

using energy conservation ,

kx^2/2 + mv^2/2 = kA^2/2

42 x 0.03^2/2 + 0.017 x 3.47^2 /2   = 42A^2 /2

xm = A = 0.0760 m = 76 mm

at t= 0

30 = 76 cos(w x 0 + phi)

cosphi = 30/76 = 66.75 degrees =1.16 rad

E = KA^2/2 = 0.121 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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