A0.23 kg mass at the end of a spring oscillates 2.0 times per second with an amplitude of 0.15 m

part a

determine the speed when it passes the equilibrium point.

part b

determine the speed when it is 0.10 m from equilibrium.

part c

determine the total energy of the system.

part d

determine the equation describing the motion of the mass, assuming that at t=0, x was a maximum.

determine the equation describing the motion of the mass, assuming that at , was a maximum.

x(t)=(0.075m)cos[2π(2.0hz)t]
x(t)=(0.15m)sin[2π(2.0hz)t]
x(t)=(0.15m)cos[2π(2.0hz)t]
x(t)=(0.15m)cos[(2.0hz)t]

A) v = 1.885 m/s

B) v = 0.39 m/s

C) E = 0.03 J

D)

Explanation:

Part A

We will use the conservation of energy to find the speed at equilibrium.

where and

Therefore,

Part B

The conservation of energy will be used again.

where

Therefore, v = 0.39 m/s.

Part C

Total energy of the system is equal to the potential energy at amplitude.

Part D

The general equation of motion in simple harmonic motion is

where is the phase angle to be determined by the initial conditions. In this case, the initial condition is that at t = 0, x is maximum. Therefore,

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.