An object relocating along the x-axis is claimed to exhibition simple harmonic motion if its position as a function of time varies as

x(t) = x0 + A cos(ωt + φ).

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The object oscillates around the equilibrium position x0. If we pick the origin of ours coordinate device such that x0 = 0, then the displacement x native the equilibrium place as a function of time is given by

x(t) = A cos(ωt + φ).

A is the amplitude that the oscillation, i.e. The maximum displacement of the thing from equilibrium, one of two people in the confident or an adverse x-direction. Basic harmonic motion is repetitive. The periodT is the moment it takes the object to finish one oscillation and return to the beginning position. Theangular frequency ω is offered by ω = 2π/T. The angular frequency is measure in radians per second. The station of the period is the frequency f = 1/T. The frequency f = 1/T = ω/2π of the motion gives the variety of complete oscillations per unit time. The is measure up in units of Hertz, (1 Hz = 1/s).

The velocity of the object as a function of time is given by

v(t) = -ω A sin(ωt + φ),

and the acceleration is provided by

a(t) = -ω2A cos(ωt + φ) = -ω2x.

The amount φ is called the phase constant. The is established by the initial conditions of the motion. If at t = 0 the object has actually its preferably displacement in the confident x-direction, then φ = 0, if it has actually its best displacement in the negative x-direction, then φ = π. If at t = 0 the fragment is moving through that equilibrium place with preferably velocity in the an unfavorable x-direction then φ = π/2. The quantity ωt + φ is referred to as the phase.

In the figure below position and velocity room plotted together a function of time for oscillatory motion with a period of 5 s. The amplitude and also the best velocity have actually arbitrary units. Position and velocity space out that phase. The velocity is zero at maximum displacement, and the displacement is zero at maximum speed.

For basic harmonic motion, the acceleration a = -ω2x is proportional to the displacement, however in the opposite direction. Simple harmonic movement is increased motion. If an item exhibits straightforward harmonic motion, a force must be acting on the object. The force is

F = ma = -mω2x.

It obeys Hooke"s law, F = -kx, with k = mω2.

The pressure exerted through a feather obeys Hooke"s law. Assume that things is attached to a spring, which is extended or compressed. Climate the feather exerts a force on the object. This force is proportional come the displacement x that the spring from that equilibrium position and also is in a direction opposite come the displacement.

F = -kx

Assume the feather is stretched a street A indigenous its equilibrium position and then released. The thing attached come the spring increases as it moves back towards the equilibrium position.

a = -(k/m)x

It gains speed as the moves in the direction of the equilibrium position since its acceleration is in the direction that its velocity. Once it is at the equilibrium position, the acceleration is zero, but the object has maximum speed. That overshoots the equilibrium position and also starts slow down, because the acceleration is now in a direction opposite come the direction the its velocity. Neglecting friction, it pertains to a stop when the spring is compressed by a distance A and also then speeds up back in the direction of the equilibrium position. That again overshoots and also comes come a protect against at the early stage position when the feather is extended a street A. The motion repeats. The object oscillates ago and forth. That executes basic harmonic motion. The angular frequency that the movement is

ω = √(k/m),

the duration is

T = 2π√(m/k),

and the frequency is

f = (1/(2π))√(k/m).

### Summary:

If the only pressure acting on an item with massive m is a Hooke"s legislation force,F = -kxthen the motion of the object is basic harmonic motion. V x gift the displacement indigenous equilibrium us have

x(t) = Acos(ωt + φ),v(t) = -ωAsin(ωt + φ),a(t) = -ω2Acos(ωt + φ) = -ω2x.ω = (k/m)½ = 2πf = 2π/T.

A = amplitudeω = angular frequencyf = frequencyT = periodφ = phase constant

Problem:

A bit oscillates with an easy harmonic motion, so the its displacement varies according come the expression x = (5 cm)cos(2t + π/6) where x is in centimeters and t is in seconds. At t = 0 find(a) the displacement that the particle,(b) that is velocity, and(c) that is acceleration.(d) uncover the period and amplitude the the motion.

Solution:

Reasoning:Analyze basic harmonic motion.x(t) = A cos(ωt + φ). A = amplitude, ω = angular frequency, φ = step constant.v(t) = -ω A sin(ωt + φ), a(t) = -ω2A cos(ωt + φ) = -ω2x.Details of the calculation:(a) The displacement as a duty of time is x(t) = Acos(ωt + φ). Below ω = 2/s, φ = π/6, and A = 5 cm. The displacement in ~ t = 0 is x(0) = (5 cm)cos(π/6) = 4.33 cm.(b) The velocity at t = 0 is v(0) = -ω(5 cm)sin(π/6) = -5 cm/s.(c) The acceleration at t = 0 is a(0) = -ω2(5 cm)cos(π/6) = -17.3 cm/s2.(d) The duration of the motion is T = 2π/ω = π s, and also the amplitude is 5 cm.Problem:

A 20 g fragment moves in simple harmonic activity with a frequency the 3 oscillations per second and an amplitude that 5 cm.(a) v what total distance go the bit move throughout one cycle of that motion?(b) What is its maximum speed? where does that occur?(c) find the maximum acceleration that the particle. Where in the activity does the preferably acceleration occur?

Solution:

Reasoning:Analyze basic harmonic motion, x(t) = A cos(ωt + φ).Details the the calculation:(a) The full distance d the fragment moves during one bicycle is from x = -A come x = +A and ago to x = -A, therefore d = 4A = 20 cm.(b) The maximum rate of the particle is vmax = ωA = 2πfA = 2π 15 cm/s = 0.94 m/s. The particle has actually maximum speed when it passes with the equilibrium position.(c) The best acceleration that the particle is amax = ω2A = (2πf)2A = 17.8 m/s2.The particle has maximum acceleration at the transforming points, where it has maximum displacement.

Assume a mass suspended indigenous a vertical spring of spring consistent k. In equilibrium the spring is extended a street x0 = mg/k. If the massive is displaced indigenous equilibrium position downward and also the feather is stretched secondary distance x, climate the full force ~ above the massive is mg - k(x0+ x) = -kx directed towards the equilibrium position. If the fixed is displaced increase by a street x, climate the full force top top the massive is mg - k(x0- x) = kx, directed towards the equilibrium position. The mass will execute basic harmonic motion. The angular frequency ω = SQRT(k/m) is the exact same for the mass oscillating on the spring in a vertical or horizontal position. But the equilibrium length of the spring about which that oscillates is different for the vertical position and the horizontal position.

Assume things attached come a spring exhibits simple harmonic motion. Allow one end of the spring be attached come a wall and allow the object relocate horizontally top top a frictionless table.

### What is the full energy that the object?

The object"s kinetic energy is

K = ½mv2 = ½mω2A2sin2(ωt + φ).

Its potential energy is elastic potential energy. The elastic potential power stored in a feather displaced a distance x indigenous its equilibrium position is U = ½kx2. The object"s potential energy as such is

U = ½kx2 = ½mω2x2 = ½mω2A2cos2(ωt + φ).

The full mechanical energy of the object is

E = K + U = ½mω2A2(sin2(ωt + φ) + cos2(ωt + φ)) = ½mω2A2.

The power E in the device is proportional come the square the the amplitude.

E = ½kA2.

It is a continuously changing mixture the kinetic energy and potential energy.

For any object executing simple harmonic activity with angular frequency ω, the restoring force F = -mω2x obeys Hooke"s law, and also therefore is a conservative force. We can define a potential energy U = ½mω2x2, and the total energy that the thing is offered by E = ½mω2A2. Due to the fact that vmax = ωA, we can likewise write E = ½mvmax2.

Problem:

A bit that hangs indigenous a spring oscillates through an angular frequency that 2 rad/s. The spring is suspended from the ceiling of an elevator car and also hangs motionless (relative come the car) together the car descends in ~ a continuous speed the 1.5 m/s. The car then all of sudden stops. Neglect the fixed of the spring.With what amplitude walk the bit oscillate?

Solution:

Reasoning:When travel in the elevator at continuous speed, the complete force ~ above the mass is zero. The pressure exerted by the spring is equal in size to the gravitational force on the mass, the spring has actually the equilibrium length of a upright spring. When the elevator all of sudden stops, the end of the feather attached come the ceiling stops. The mass, yet has momentum, p = mv, and also therefore starts extending the spring. That moves v the equilibrium place of the upright spring with its preferably velocity vmax= 1.5 m/s.Its velocity together a duty of time is v(t) = -ωAsin(ωt + φ). Details of the calculation:Since vmax = ωA and ω = 2/s, the amplitude the the amplitude that the oscillations is A = 0.75 m.Problem:

A mass-spring system oscillates v an amplitude that 3.5 cm. If the force constant of the feather of 250 N/m and the massive is 0.5 kg, determine(a) the mechanical power of the system,(b) the maximum rate of the mass, and(c) the best acceleration.

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Solution:

Reasoning:The mechanical energy of a mechanism executing simple harmonic movement is E = ½kA2 = ½mω2A2.Details that the calculation:(a) We have m = 0.5 kg, A = 0.035 m, k = 250 N/m, ω2 = k/m = 500/s2, ω = 22.36/s.The mechanical power of the mechanism is E = ½kA2 = 0.153 J.(b) The maximum rate of the fixed is vmax = ωA = 0.78 m/s.(c) The preferably acceleration is amax = ω2A = 17.5 m/s2.