Question

A simple harmonic oscillator is made up of a mass-spring system, with mass of 1.31 kg and a spring constant k = 148 N/m. At time t=1.66 s, the position and velocity of the block are x = 0.135 m and v = 3.958 m/s. What is the amplitude, x_m, of the oscillations?

Your answer should be in m, but enter only the numerical part in the box.

Answer 1

Considering the The system as frictionless ,total mechanical energy of the system is the sum of kinetic energy and spring potential energy.

Mechanical energy = Kinetic energy + potential energy

M.E = (1/2)*m*v² + (1/2)*k*x²

using the known values at t = 1.66 seconds,

M.E = (1/2)*1.31*3.958² + (1/2)*148*0.135²

M.E = 10.261 + 1.348

M.E = 11.609 J

Thus, total mechanical energy of the system is 11.609 J.

Now at the longest elongation(amplitude) of the spring the whole kinetic energy is converted into spring potential energy.so,whole mechanical energy only consist of spring potential energy.

M.E = Spring potential energy (x=A)

11.609 = (1/2)*k*A²

where A is the amplitude.

11.609 = (1/2)*148*A²

solving for A,

A = 0.396 m

amplitude of these oscillations is 0.396 m.