Question

An oscillator consists of a block attached to a spring (k = 483 N/m). At some time t, the position (measured from the system's equilibrium location), velocity, and acceleration of the block are x = 0.0632 m, v = -18.4 m/s, and a = -105 m/s2. Calculate (a) the frequency of oscillation, (b) the mass of the block, and (c) the amplitude of the motion.

Answer 1

Given

Displacement x = 0.0632 m

Velocity v = -18.4 m/s

Acceleration a = -105 m/s²

Spring constant k = 483 N/m

Solution

A)

Acceleration

a = -ω² x

ω² = -a/x

ω² = -105/-0.0632

ω² = 1661.39

ω = 40.76 rad/s

Frequency

F = ω/2π

F = 40.76 / 2 x 3.14

F = 6.49 Hz

B)

ω = √(k/m)

ω² = k/m

105/0.0632 = 483 /m

m = 483 x 0.0632 / 105

m = 0.29072 kg

m = 0.291 kg

C)

At any point the total energy of the block

Kinetic energy + elastic potential energy

Total energy at this instance

E = E_k + E_p

E = ½ mv² + ½ kx²

E = (mv² + kx²)/2

E = (0.29072 x 18.4²  + 483 x 0.0632² )/2

E = 50.2 J

When the displacement is at its maximum all kinetic energy will be converted into potential energy

At amplitude A ( or maximum displacement)

Total energy

E = E_k + E_p

E = 0 + ½ kA²

50.2 = ½ x 483 x A²

A = 0.4558 m