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Wize AP Physics C: Mechanics Textbook > Unit 6: Oscillations (6-14%)

Velocity and Acceleration

Basic Concepts of Oscillation Previous Section
SHM as the Projection of a Uniform Circular MorionNext Section
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Velocity and Acceleration in SHM


Displacement of an oscillator respect to the equilibrium position is found by:

x(t)=Acos⁡(ωt+φ)x\left(t\right)=A\cos\left(\omega t+\varphi\right)



Velocity

We can find the velocity of an oscillator by taking derivate of displacement respect to time:

v(t)=dxdt=−Aωsin⁡(ωt+φ)\boxed{v\left(t\right)=\frac{dx}{dt}=-A\omega\sin\left(\omega t+\varphi\right)}







Wize Concept
  • Velocity of an oscillator is an oscillating function with the same period as x(t)x(t).
  • Velocity is changing in the following range:
−Aω≤v(t)≤+Aω-A\omega\le v(t)\le +A\omega


Watch Out!
Velocity is a vector and it could be both positive or negative but speed is the magnitude of velocity vector and it is always positive! Hence:
Minimum speed=0\text{Minimum speed}=0



Acceleration

We can find the acceleration of an oscillator by taking derivate of velocity respect to time:

a(t)=dvdt=−Aω2cos⁡(ωt+φ)\boxed{a\left(t\right)=\frac{dv}{dt}=-A\omega^2\cos\left(\omega t+\varphi\right)}








Wize Concept
  • Acceleration of an oscillator is an oscillating function with the same period as x(t)x(t) and v(t)v(t).
  • Acceleration is changing in the following range:
−Aω2≤a(t)≤+Aω2-A\omega^2\le a(t)\le +A\omega^2



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Velocity and Acceleration in SHM


The position as a function of time for an object undergoing SHM is usually described by a cosine function.


x(t) = Acos⁡ωtx\left(t\right)\ =\ A\cos\omega t


Velocity

We can find the velocity of an oscillator by taking derivate of displacement respect to time:

v(t)=dxdt=−Aωsin⁡(ωt)\boxed{v\left(t\right)=\frac{dx}{dt}=-A\omega\sin\left(\omega t\right)}


Wize Concept
  • Velocity of an oscillator is an oscillating function with the same period as x(t)x(t).
  • Velocity is changing in the following range:
−Aω≤v(t)≤+Aω-A\omega\le v(t)\le +A\omega


Watch Out!
Velocity is a vector and it could be both positive or negative but speed is the magnitude of velocity vector and it is always positive! Hence:
Minimum speed=0\text{Minimum speed}=0



Acceleration

We can find the acceleration of an oscillator by taking derivate of velocity respect to time:

a(t)=dvdt=−Aω2cos⁡(ωt)\boxed{a\left(t\right)=\frac{dv}{dt}=-A\omega^2\cos\left(\omega t\right)}


Wize Concept
  • Acceleration of an oscillator is an oscillating function with the same period as x(t)x(t) and v(t)v(t).
  • Acceleration is changing in the following range:
−Aω2≤a(t)≤+Aω2-A\omega^2\le a(t)\le +A\omega^2



Finding Acceleration by Knowing Displacement

Let's look at the equations for displacement and acceleration of an oscillator one more time:

x(t)=Acos⁡(ωt)a(t)=−Aω2cos⁡(ωt)\begin{array}{c}x\left(t\right)=A\cos\left(\omega t\right)\\\\a\left(t\right)=-A\omega^2\cos\left(\omega t\right) \end{array}

By comparing displacement and acceleration functions we can find the following nice relationship between them:

a(t)=−ω2x(t)\boxed{a(t)=-\omega^2x(t)}

Wize Tip
Note that based on the above equation, acceleration is maximum when displacement is minimum and vice versa.

Exam Tip
Above equation could be used to find acceleration of an oscillator at any moment if you know its position!


Relation Between ω\colorOne{\omega }and k\colorOne{k}

We just learned that any physical system with a force in the form of F=−kxF=-kxundergoes a simple harmonic motion. Here kkis a constant which depends on the physical properties of the system.

We can fins find a relationship between this constant (k)(k)and the angular frequency of oscillation (ω)(\omega)by looking at the second law of Newton:


F=ma=−kx→a=−kmx=−ω2x→km=ω2 →ω=km\begin{array}{c}F=ma=-kx\to a=-\dfrac{k}{m}x=-\omega^2x\\ \\\to\dfrac{k}{m}=\omega^2\ \rightarrow \\\\ \boxed{\omega=\sqrt{\dfrac{k}{m}}}\end{array}






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Finding Acceleration by Knowing Displacement


As we discussed before, here are the equations for displacement and acceleration of an oscillator:

x(t)=Acos⁡(ωt+φ)a(t)=−Aω2cos⁡(ωt+φ)\begin{array}{c}x\left(t\right)=A\cos\left(\omega t+\varphi\right)\\\\a\left(t\right)=-A\omega^2\cos\left(\omega t+\varphi\right) \end{array}

By comparing displacement and acceleration functions we can find the following nice relationship between them:

a(t)=−ω2x(t)\boxed{a(t)=-\omega^2x(t)}

Wize Tip
Note that based on the above equation, acceleration is maximum when displacement is minimum and vice versa.

Exam Tip
Above equation could be used to find acceleration of an oscillator at any moment if you know its position!


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Relation Between ω\colorOne{\omega }and k\colorOne{k}

We just learned that any physical system with a force in the form of F=−kxF=-kxundergoes a simple harmonic motion. Here kkis a constant which depends on the physical properties of the system.

We can fins find a relationship between this constant (k)(k)and the angular frequency of oscillation (ω)(\omega)by looking at the second law of Newton:


F=ma=−kx→a=−kmx=−ω2x→km=ω2 →ω=km\begin{array}{c}F=ma=-kx\to a=-\dfrac{k}{m}x=-\omega^2x\\ \\\to\dfrac{k}{m}=\omega^2\ \rightarrow \\\\ \boxed{\omega=\sqrt{\dfrac{k}{m}}}\end{array}





A 12.5 kg mass is hung on a spring with spring constant 80 N/m. It stretches a distance of 2.1 m and released. Its phase shift is 1.57. What is the instantaneous velocity at 3 seconds?


m = 12.5 kg
A = 2.1 m
phi = 1.57
t = 3 s

find w:
w = sqrt (k/m) = sqrt (80/12.5) = 2.5 Hz.

v(t) = A w cos(wt + phi)
v(3) = 2.1(2.5)(cos((2.5)(3s)) + 1.57)
= 5.25 cos(9.07)
= -4.9 m/s.
The position of an oscillating mass in a simple pendulum is given as: x(t)=3.56cos⁡(12t+π)x(t)=3.56\cos(12t+\pi).
What is the maximum velocity of the mass?

Practice: Getting Info About Acceleration by Looking at Displacement


A harmonic oscillator's motion is described by the following graph
Which of the following statement is correct about the acceleration (NOT the magnitude) of the oscillator?