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Physics 2210 Fall 2015
smartPhysics 21 Simple Harmonic Motion
22 Simple and Physical Pendulum 11/25/2015
Exam 4: smartPhysics units 14-20 Midterm Exam 4: Day: Fri Dec. 04, 2015 Time: regular class time Section 01 12:55 – 1:45 pm Section 10 02:00 – 02:50 pm
Location: FMAB 15 practice problems for Exam 4 posted on CANVAS and to the Class Web Page (http://www.physics.utah.edu/~woolf/2210_Jui/rev4.pdf)
Derivation Already Given in Pre-lecture 1/4
In each case we have the usual relationship between force (exerted by the spring on the block) and the position (displacement from equilibrium) of the block:
�⃗� = −𝑘𝑘𝚤̂ Applying Newton’s 2nd Law then we have (noting that 𝑎𝑥 = 𝑑2𝑘 𝑑𝑑2⁄ ):
𝑑2𝑘𝑑𝑑2 = −
𝑘𝑚𝑘
This is an equation that involves the 2nd derivative of position, 𝒙, with respect to time, t
𝐿 = 𝐿0 − 𝑏 𝑚
𝑘 = −𝑏
𝐿 = 𝐿0 𝑚
𝑘 = 0
𝐿 = 𝐿0 + 𝑏 𝑚
𝑘 = +𝑏
�⃗� = +𝑘𝑏𝚤̂ 𝑏 is a positive length
�⃗� = 0
�⃗� = −𝑘𝑏𝚤̂
Equilibrium Position
Derivation Already Given in Pre-lecture 2/4
The solution to the equation is equivalent to integrating twice: in other words, it will have TWO “integration” constants that are to be determined by the particular conditions But first: let’s guess at the solution we can try 𝑘1 = 𝑐1 sin𝜔𝑑 :
𝑑𝑘1𝑑𝑑 = 𝜔𝑐1 cos𝜔𝑑 →
𝑑2𝑘1𝑑𝑑2 =
𝑑𝑑𝑑
𝑑𝑘1𝑑𝑑 =
𝑑𝑑𝑑
𝜔𝑐1 cos𝜔𝑑 = −𝜔 ∙ 𝜔𝑐1 cos𝜔𝑑
= −𝜔2𝑐1 cos𝜔𝑑 But the equation requires:
𝑑2𝑘1𝑑𝑑2 = −
𝑘𝑚𝑘1 → −𝜔2𝑐1 cos𝜔𝑑 = −
𝑘𝑚 𝑐1 cos𝜔𝑑
This means 𝑘1 = 𝑐1 sin𝜔𝑑 can be a solution provided we satisfy the condition:
𝜔2 =𝑘𝑚 → 𝜔 = 𝑘 𝑚⁄
Unit check:
𝑘 𝑚⁄ =N m⁄
kg =kg ∙ m
s2 ∙1m ∙
1kg = s−2 = s−1
This is what we need: because the argument of a sine/cosine function needs to be unitless.
𝐿 = 𝐿0 + 𝑘 𝑚
𝑘
�⃗� = −𝑘𝑘𝚤̂
Derivation Already Given in Pre-lecture 3/4
We can also try 𝑘2 = 𝑐2 cos𝜔𝑑 : 𝑑𝑘2𝑑𝑑 = −𝜔𝑐2 sin𝜔𝑑 →
𝑑2𝑘2𝑑𝑑2 =
𝑑𝑑𝑑
𝑑𝑘2𝑑𝑑 =
𝑑𝑑𝑑
−𝜔𝑐2 sin𝜔𝑑 = 𝜔 ∙ −𝜔𝑐2 cos𝜔𝑑
= −𝜔2𝑐2 cos𝜔𝑑 But the equation requires:
𝑑2𝑘2𝑑𝑑2 = −
𝑘𝑚𝑘2 → −𝜔2𝑐2 cos𝜔𝑑 = −
𝑘𝑚 𝑐2 cos𝜔𝑑
This means 𝑘2 = 𝑐2 cos𝜔𝑑 can ALSO be a solution provided we satisfy the SAME condition:
𝜔2 =𝑘𝑚 → 𝜔 = 𝑘 𝑚⁄
NOTE: The equation involves only the function 𝒙(𝒕), and its 2nd derivative �̈� ≡ 𝒅𝟐𝒙 𝒅𝒕𝟐⁄ , BUT NOT its powers (like 𝒙𝟐, or �̈�𝟐). We say that the equation is LINEAR This means any linear combination of the two solutions is also a solution
𝑘(𝑑) = 𝑐1 sin𝜔𝑑 + 𝑐2 cos𝜔𝑑 is also a solution for 𝜔 = 𝑘 𝑚⁄ Since this form involves TWO integration constants, 𝑐1 and 𝑐2, it is a GENERAL solution Another general solution (smartPhysics is inconsistent) has the form 𝑘 𝑑 = 𝐴 cos 𝜔𝑑 + 𝜙
𝐿 = 𝐿0 + 𝑘 𝑚
𝑘
�⃗� = −𝑘𝑘𝚤̂
Derivation Already Given in Pre-lecture 4/4
For 𝜔 = 𝑘 𝑚⁄ , 𝑘(𝑑) = 𝑐1 sin𝜔𝑑 + 𝑐2 cos𝜔𝑑 and 𝑘 𝑑 = 𝐴 cos 𝜔𝑑 + 𝜙 are equivalent : Trigonometric identity
sin 𝑢 + 𝑣 = sin𝑢 cos𝑣 + sin 𝑣 cos𝑢 cos 𝑢 + 𝑣 = cos𝑢 cos𝑣 − sin𝑣 sin𝑢
So we can apply the second to
𝐴 cos 𝜔𝑑 + 𝜙 = 𝐴 cos𝜔𝑑 cos𝜙 − 𝐴 sin𝜙 sin𝜔𝑑
Now recalling that sine is an odd function, and cosine an even function
𝐴 cos 𝜔𝑑 + 𝜙 = − 𝐴 sin𝜙 sin𝜔𝑑 + 𝐴 cos𝜙 cos𝜔𝑑
In other words to make the two forms be equivalent, we identify
𝑐1 = −𝐴 sin𝜙 , 𝑐2 = 𝐴 cos𝜙
Or conversely
𝐴 = 𝑐12 + 𝑐22, 𝜙 = tan−1−𝑐1𝑐2
𝐿 = 𝐿0 + 𝑘 𝑚
𝑘
�⃗� = −𝑘𝑘𝚤̂
Unit of angular frequency 𝜔: s−1 or rad/s*** To complete one period of the motion, the argument to the cosine/sine must advance by 2𝜋 i.e. the period, T, is given by 𝜔𝑇 = 2𝜋 → 𝑇 = 2𝜋/𝜔 (unit of period: seconds) The “frequency” (number of complete cycles per unit time) is given by 𝑓 = 1 𝑇⁄ = 𝜔/2𝜋 Unit of frequency: s−1 ≡ ℎ𝑒𝑒𝑑𝑒 (Hz) The maximum displacement from equilibrium is 𝐴, known as the amplitude (unit: meters) Maximum speed: 𝑣𝑚𝑚𝑥 = 𝜔𝐴, maximum acceleration: 𝑎𝑚𝑚𝑥 = 𝜔2𝐴
Uniform Circular Motion Model of Simple Harmonic Motion
Top: Particle in uniform circular motion with radius 𝑒 = 𝐴 and constant angular velocity 𝜔 and initial angle 𝜙
Its Cartesian coordinates are: 𝑘 = 𝐴 cos 𝜔𝑑 + 𝜙 𝑦 = 𝐴 sin 𝜔𝑑 + 𝜙
In this context, 𝜙 = 𝜃 𝑑 = 0 is the angular position of the particle at 𝑑 = 0
Bottom: Harmonic oscillator with amplitude 𝑨, frequency 𝒇 = 𝝎 𝟐𝝅⁄ , initial phase angle 𝝓. i.e at time 𝑑 = 0:
𝑘 = 𝑘0 = 𝐴 cos𝜙 𝑣 = 𝑣0 = −𝜔𝐴 sin𝜙
𝑘
𝑦
𝐴 𝜃 = 𝜙 + 𝜔𝑑
Equilibrium at 𝑘 = 0 𝑚
𝑘 = 𝐴 cos 𝜔𝑑 + 𝜙
Poll 11-25-01
A mass on a spring moves with simple harmonic motion as shown.
Where is the acceleration of the mass most positive?
A. x = -A B. x = 0 C. x = +A
Poll 11-25-02
In the two cases shown the mass and the spring are identical but the amplitude of the simple harmonic motion is twice as big in Case 2 as in Case 1.
How are the maximum velocities in the two cases related?
A. Vmax,2 = Vmax,1 B. Vmax,2 = 2 Vmax,1 C. Vmax,2 = 4 Vmax,1
Demo/Example 21.1 1. Hang vertically from pivot, relaxed 2. With 500 g mass attached: the spring is
stretched from relaxed length to the NEW equilibrium
3. With extra 100 g attached, stretches ∆𝐿 from relaxed length
Calculate Spring Constant:
𝑘 ≡∆𝐹∆𝐿
=0.100 kg ∙ 9.8 m s2⁄
𝟎.𝟏𝟎 m = 9.8 N m⁄
If we just hang a 500 g mass, and take configuration (2) to be equilibrium Then for displacement 𝑘 about the equilibrium configuration of (2) we have
𝜔 = 𝑘 𝑚⁄ = 9.8 N m⁄ 0.50 kg⁄= 𝟒.𝟒𝟒 s−1
The period is predicted to be
𝑇 =2𝜋𝜔 =
2𝜋𝟒.𝟒𝟒s−1 = 𝟏.𝟒𝟐 S
Measurement: time for 10 cycles: 14.7 s Measured period = 1.47 s
500 g
500 g
100 g
∆𝐿
(1)
(2)
(3)
𝑘
𝐿0
EQULIBRIUM
Demo/Example 22.1 Simple Pendulum
This is the rotational (approximate) equivalent of a simple harmonic oscillator The moment of inertia about the pivot for a small particle is
𝐼 = 𝑚𝑒2 = 𝑚𝐿2 The torque exerted by the force of gravity is
𝜏 = 𝑒𝐹 sin𝜙𝑟𝑟 = 𝐿 ∙ 𝑚𝑚 ∙ sin −𝜃 = −𝑚𝑚𝐿 sin𝜃 Where we have observed that SINE is an odd function: sin −𝜃 = − sin𝜃, so the equation of motion is:
𝑑2𝜃𝑑𝑑2 =
𝜏𝐼 =
−𝑚𝑚𝐿 sin𝜃𝑚𝐿2 = −
𝑚𝐿 sin𝜃 ≈ −
𝑚𝐿 𝜃
𝜃
P
𝑚𝑚 −𝜃
𝑚
𝐿
Which has the form
𝑑2𝜃𝑑𝑑2 = −𝜔2𝜃, 𝜔 =
𝑚𝐿 =
9.8 m s2⁄𝟎.𝟓𝟎m = 𝟒.𝟒𝟒 s−1
And the period is related to the angular velocity/frequency by (period is the time it takes to do one cycle: i.e. 𝜔𝑇 = 2𝜋
𝑇 =2𝜋𝜔 =
2𝜋𝟒.𝟒𝟒s−1 = 𝟏.𝟒𝟐s
Measured Period: 14.3s/10 = 1.43 s
