Question

1.

A 0.12kg body is connected to a wall by a spring with a spring constant of 570 N/m. The body experiences simple oscillatory motion when pulled from its equilibrium rightward by 0.080m and then released from rest. what is the displacement of the block after 0.20s.

2.

An object connected to a spring (with a spring constant of 29.8 N/m) is displaced 0.232 meter from equilibrium on a frictionless horizontal tabletop; upon release, the object experiences simple harmonic motion as it oscillates horizontally with ω = 2.81 radians per second. what is the potential energy stored in the mass-spring system after 1.42 seconds.

Answer 1

1)

Given m = 0.12 kg

spring constant K = 570 N/m

Amplitude A = 0.08 m

t = 0.2 s

Angular frequency   omega = sqrt(k/m)

omega = sqrt(570/0.12)

omega = 68.92 rad/s

The displacement is given by

x = A * cos(omega * t)

x = 0.08 * cos(68.92 * 0.2)

x = 0.0276 m

2)

K = 29.8 N/m

Amplitude A = 0.232 m

omega = 2.81 rad/s

time t = 1.42 m

The displacement is given by

x = A * cos(omega * t)

x = 0.232 * cos(2.81 * 1.42)

x = - 0.153 m

The potential energy is

U = 1/2 * k * x^2

U = 1/2 * 29.8 * (-0.153)^2

U = 0.35 J