The equation of motion of a simple harmonic oscillator is given by
x(t) = (9.0 cm)cos(12πt) − (5.4 cm)sin(12πt),
where t is in seconds.
(a) Find the amplitude.
Determine the period.
(c) Determine the initial phase.
To find the amplitude of the simple harmonic oscillator, we need to consider the general equation of motion for a simple harmonic oscillator:
x(t) = A * cos(ωt + ϕ),
where:
x(t) represents the displacement at time t,
A is the amplitude (maximum displacement) of the oscillator,
ω is the angular frequency,
t represents time, and
ϕ is the phase constant.
Comparing this general equation with the given equation, we can conclude:
A = √{[cos(12πt)]² + [sin(12πt)]²}.
Simplifying this expression, we get:
A = √{cos²(12πt) + sin²(12πt)}
Using the trigonometric identity cos²θ + sin²θ = 1, we can further simplify:
A = √1 = 1.
The amplitude of the given simple harmonic oscillator is 1.
To find the period, we need to use the formula for the period of a simple harmonic oscillator:
T = 2π/ω,
where T represents the period and ω is the angular frequency.
Comparing this equation with the given equation, we can conclude:
ω = 12π.
Substituting this value into the formula for the period, we get:
T = 2π/(12π) = 1/6 s.
The period of the given simple harmonic oscillator is 1/6 s, or approximately 0.167 s.
To determine the initial phase, we need to consider the general equation of motion for a simple harmonic oscillator:
x(t) = A * cos(ωt + ϕ),
where ϕ represents the initial phase.
Comparing this general equation with the given equation, we can conclude:
ϕ = arctan(-5.4 cm / 9.0 cm).
Using a calculator or trigonometric tables, we can find the value of this expression:
ϕ ≈ -30.96°.
Therefore, the initial phase of the given simple harmonic oscillator is approximately -30.96°.