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Physics Lesson 11.2.3 - Speed and Acceleration of Oscillating Particles in a Wave

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Welcome to our Physics lesson on Speed and Acceleration of Oscillating Particles in a Wave, this is the third lesson of our suite of physics lessons covering the topic of General Equation of Waves, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Speed and Acceleration of Oscillating Particles in a Wave

Just like in SHM, we can find the speed of oscillating particles in a wave by taking the first derivative of y-position with respect to the time. Be careful not confuse the oscillating speed of particles (it lies in the y-direction and is not uniform) with the wave speed (which is horizontal and uniform). Thus,

vy(t) = dy/dt
= d[ymax × sin(k × x - ω × t + φ) ]/dt
= ymax × ω × cos(k × x - ω × t + φ)

Likewise, we obtain for the vertical acceleration of oscillating particles,

ay (t) = dv/dt
= -d[ymax × ω × cos(k × x - ω × t + φ) ]/dt
= -ymax × ω2 × sin(k × x - ω × t + φ)
= -ω2 × y(t)

Example 2

A wave oscillates according the equation

y(x,t) = 3 × sin(6π × x - 3π × t)

where x and y are in metres and t is in seconds. Calculate:

  1. Wavelength
  2. Period of the wave
  3. Wave's speed
  4. Vertical position at t = 2 s of a point of wave that oscillates according the vertical line x = 3 m
  5. Vertical speed of the above point at t = 2 s

Solution 2

a) From the general equation of waves

y(x,t) = ymax × sin(k × x - ω × t + φ)

And comparing it with the actual equation

y(x,t) = 3 × sin(6π × x - 3π × t)

we extract the following values:

ymax = A = 3 m
k = 6π rad/m
ω = 3π rad/s

and

φ = 0

Thus, giving that

k = 2π/λ = 6π

we obtain for the wavelength λ

λ = 2π/6π = 1/3 = 0.33 m

b) Since

ω = 2π/T = 3π

we obtain for the period

T = 2π/3π = 0.67 s

c) Now we can use the simplified equation of wave

v = λ × f = λ/T

to find the wave speed. Thus,

v = 0.33 m/0.67 s = 0.5 m/s

d) Since the vertical position y of a point of wave at any instant t is given by the equation

y(x,t) = ymax × sin(k × x - ω × t + φ)

and since in the specific case this equation is

y(x,t) = 3 × sin(6π × x - 3π × t)

we obtain for x = 3 and t = 2

y(3m,2s) = 3 × sin(6π × 3 - 3π × 2)
= 3 × sin12π
= 0

This result means the given point of the wave is at the equilibrium position at t = 2 s.

e) Vertical speed is the first derivative of vertical position in respect to the time. Thus, we have

vy = dy/dt
= d[3 × sin(6π × x - 3π × t) ]/dt
= 3 × 6π × cos(6π × x - 3π × t)
= 18π × cos(6π × x - 3π × t)

Thus, for x = 3 and t = 2 we obtain for the vertical speed vy

vy (3m,2s) = 18π × cos(6π × 3 - 3π × 2)
= 18π × cos(12π)
= 18 × 3.14 × 1
= 56.52 m/s

You have reached the end of Physics lesson 11.2.3 Speed and Acceleration of Oscillating Particles in a Wave. There are 3 lessons in this physics tutorial covering General Equation of Waves, you can access all the lessons from this tutorial below.

More General Equation of Waves Lessons and Learning Resources

Waves Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial		Revision
Notes		Revision
Questions
11.2General Equation of Waves
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
11.2.1Wave Function. General Equation of Waves
11.2.2How to Obtain the Simplified Wave Equation from the General Equation of Waves
11.2.3Speed and Acceleration of Oscillating Particles in a Wave

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  4. Waves Practice Questions: General Equation of Waves. Test and improve your knowledge of General Equation of Waves with example questins and answers
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