Worked examples · Sheet 1

Equations of motion of harmonic motion

Theory used: Simple harmonic motion Periodic motion, period and frequency Velocity and acceleration in harmonic motion Velocity as a function of position Amplitude and phase from the initial conditions

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A particle moves along the $x$ -axis with equation of motion

x (t) = 0.050 cos (4 π t + \frac{π}{3}),

with $x$ in meters and $t$ in seconds. Determine the amplitude, angular frequency, period and frequency, the maximum values of the velocity and of the acceleration, and the position and velocity at the instant $t = 0$ .

A = 0.050 m, ω = 4 π rad/s, φ = \frac{π}{3}

T = \frac{2 π}{ω} = \frac{2 π}{4 π} s = 0.50 s, f = \frac{1}{T} = 2.0 Hz

v_{m a x} = A ω = 0.050 \cdot 4 π m/s = 0.628 m/s

a_{m a x} = A ω^{2} = 0.050 \cdot 16 π^{2} m/ s^{2} = 7.90 m/ s^{2}

x (0) = A cos \frac{π}{3} = 0.050 \cdot \frac{1}{2} m = 0.025 m

v (0) = - A ω sin \frac{π}{3} = - 0.628 \cdot \frac{3}{2} m/s = - 0.544 m/s

Equation of motion with

x

in centimeters and

t

in seconds: the particle starts from

x (0) = 2.5 cm

toward decreasing values and performs two oscillations per second

The equation of motion has the form $A cos (ω t + φ)$ of simple harmonic motion, so amplitude, angular frequency and initial phase can be read off directly, since the coefficient of the cosine is $A$ , the coefficient of $t$ is $ω$ and the constant term of the argument is $φ$ . The period is $\frac{2 π}{ω}$ and the frequency is its reciprocal. Differentiating the equation of motion gives $v = - A ω sin (ω t + φ)$ and $a = - A ω^{2} cos (ω t + φ)$ ; since sine and cosine have maximum value $1$ , the maximum magnitudes are $A ω$ and $A ω^{2}$ . At the instant $t = 0$ the argument equals $\frac{π}{3}$ , therefore the particle is at $x = 0.025 m$ , at half the amplitude, and has negative velocity, that is, it is moving toward the origin.

Exercise 2

A body performs simple harmonic motion. When passing through the equilibrium position its velocity has magnitude $0.60 m/s$ , and at the turning points its acceleration has magnitude $7.2 m/ s^{2}$ . Determine the angular frequency, the amplitude, the period and the magnitude of the velocity at the position $x = 0.030 m$ .

v_{m a x} = A ω, a_{m a x} = A ω^{2}

ω = \frac{a _{m a x}}{v _{m a x}} = \frac{7.2}{0.60} rad/s = 12 rad/s, A = \frac{v _{m a x}}{ω} = \frac{0.60}{12} m = 0.050 m

T = \frac{2 π}{ω} = \frac{2 π}{12} s = 0.524 s

∣ v ∣ = ω A^{2} - x^{2} = 12 0.05 0^{2} - 0.03 0^{2} m/s = 12 \cdot 0.040 m/s = 0.48 m/s

In harmonic motion the velocity is maximum when passing through equilibrium and the acceleration is maximum at the turning points $x = \pm A$ , so the two data are $A ω$ and $A ω^{2}$ . The ratio of the two expressions eliminates $A$ and gives $ω$ ; once the angular frequency is known, the amplitude is obtained from $v_{m a x} = A ω$ . For the velocity at a given position we use the relation $∣ v ∣ = ω A^{2} - x^{2}$ , which links velocity and position without going through the instant. Therefore at the position $x = 0.030 m$ the body has velocity of magnitude $0.48 m/s$ . As a dimensional check, $\frac{m/ s ^{2}}{m/s} = s^{- 1}$ , which is the unit of an angular frequency.

Exercise 3

A particle in harmonic motion with angular frequency $ω = 5.0 rad/s$ is at the instant $t = 0$ at the position $x_{0} = 0.040 m$ with velocity $v_{0} = 0.15 m/s$ . Write the equation of motion.

A = x_{0}^{2} + \frac{v _{0}^{2}}{ω ^{2}} = 0.04 0^{2} + (\frac{0.15}{5.0})^{2} m = 0.0016 + 0.0009 m = 0.050 m

⎩ ⎨ ⎧ cos φ = \frac{x _{0}}{A} = \frac{0.040}{0.050} = 0.80 sin φ = - \frac{v _{0}}{A ω} = - \frac{0.15}{0.050 \cdot 5.0} = - 0.60 ⟹ φ = - 0.644 rad

x (t) = 0.050 cos (5.0 t - 0.644) m

The equation of motion is $x (t) = A cos (ω t + φ)$ and the angular frequency is known, so $A$ and $φ$ remain to be determined from the initial conditions. Imposing $x (0) = A cos φ = x_{0}$ and $v (0) = - A ω sin φ = v_{0}$ and adding the squares gives the amplitude, which equals $0.050 m$ . The phase must satisfy the two conditions of the system together; a positive cosine and a negative sine identify an angle in the fourth quadrant, so $φ = - arccos (0.80) = - 0.644 rad$ . As a check, $x (0) = 0.050 \cdot 0.80 m = 0.040 m$ and $v (0) = - 0.050 \cdot 5.0 \cdot (- 0.60) m/s = 0.15 m/s$ , as required.

Exercise 4

A particle oscillates in harmonic motion with amplitude $A = 0.10 m$ and period $T = 1.2 s$ , and at the instant $t = 0$ it starts from rest at the position $x = A$ . Determine the first instant at which the particle passes through $x = - \frac{A}{2}$ and its velocity at that instant.

ω = \frac{2 π}{T} = \frac{2 π}{1.2} rad/s = 5.24 rad/s, x (t) = A cos (ω t)

A cos (ω t) = - \frac{A}{2} ⟺ cos (ω t) = - \frac{1}{2}, ω t_{1} = \frac{2 π}{3}

t_{1} = \frac{2 π}{3 ω} = \frac{T}{3} = 0.40 s

v (t_{1}) = - A ω sin \frac{2 π}{3} = - 0.10 \cdot 5.24 \cdot \frac{3}{2} m/s = - 0.453 m/s

The particle starts from rest at the extreme $x = A$ , so $x_{0} = A$ and $v_{0} = 0$ ; by the initial conditions the initial phase is $φ = 0$ and the equation of motion is $A cos (ω t)$ . The equation $cos (ω t) = - \frac{1}{2}$ has infinitely many solutions; for $0 \leq ω t \leq π$ the cosine decreases from $1$ to $- 1$ , therefore the first positive solution is $ω t = \frac{2 π}{3}$ , that is, one third of the period. The velocity is obtained from the derivative of the equation of motion and is negative, because in the first half of the period the particle goes from the positive extreme toward the negative one. As a check, $ω A^{2} - x^{2} = 5.24 \cdot 0.0100 - 0.0025 m/s = 0.453 m/s$ , which coincides with the magnitude found.