References:

1. Waves and Optics | Online Physics Courses | Rice University
   • Searched by wave optics course in DDG

Introduction

Examples

Solving Kinematics with Two Steps:

1. Free Body Diagram -> Newton’s Second Law -> EOM
2. Solve: Integrating or Guessing -> Initial Conditions

Solve Kinematics

Objectives:

1. Review the steps for solving a calculus-based physics problem before studying simple harmonic motion

   (2025-07-17T23:33)

   • 求一个运动学问题，本质是对时间积分。
   • 初始条件是什么？初始条件是时间 t=0 的时候。

Problems:

1. What is the trajectory of an object under a constant force?

2. What is the position $(x_t,y_t)$ of the object in an arbitrary time?

3. What is x, as a function of time? $x = f_1(t)$

   What is y, as a function of time? $y = f_2(t)$

Notes:

1. Example of firing a cannonball on a cliff

   • Physical situation illustration and Free Body Diagram:

   • Steps:

     1. Apply laws of physics: Differential Equations Of Motion (E.O.M.) in the both x-direction and y-direction

        $$ \begin{array}{c|c} \begin{aligned} ∑F_y &= m a_y \\\ -mg &= m a_y \\\ -g &= a_y \quad \text{Cancel $m$} \\\ -g &= \frac{d (\frac{dy}{dt})}{dt} \quad \text{Represent $a_y$ with y,t} \end{aligned} & \begin{aligned} ∑F_x &= m a_x \\\ 0 &= m a_x \\\ 0 &= a_x \\\ 0 &= \frac{d^2 x}{dt^2} \\\ \end{aligned} \\\ \end{array} \\\ $$

        • Newton’s second law: The sum of forces in the y-direction is the mass times the acceleration in the y.

        • There is only the gravity pulling it down after it takes off.

        • To find $y$, as a $f(t)$, the equation needs to be transformed to a form, including y and t.

          To determine y, which is expected to be a function of t, the acceleration $a_y$ should be written as a representation containing both y and t, i.e., the secondary acceleration of y w.r.t. t.

     2. Solve the differential equations by integrating them

        $$ \begin{array}{c|c} \begin{aligned} -g &= \frac{d^2 y}{dt^2} \\\ -gt + c_1 &= \frac{dy}{dt} \\\ -\frac{1}{2} gt^2 + c_1 t + c_2 &= y \end{aligned} & \begin{aligned} 0 &= \frac{d^2 x}{dt^2} \\\ c_3 & = \frac{dx}{dt} \\\ c_3 t + c_4 &= x \end{aligned} \end{array} $$

        • Those constants of the integrations: $c_1,\ c_2,\ c_3,\ c_4$ depend on the initial conditions.

          When $t = 0$, position $y = y_0$ and velocity in y $\frac{dy}{dt} = v_{0y}$. Therefore,

          $$ c_2 = y_0 \\\ c_1 = v_{0y} $$

          Similarly, in the x-direction, according to the condition at $t=0$, i.e., $x = x_0$ and $\frac{dx}{dt} = v_{0x}$, there are:

          $$ c_3 = v_{0x} \\\ c_4 = x_0 $$

        • The velocity in the y-direction is: $\frac{dy}{dt} = -gt + v_{0y}$, which is accelerating downwards. Whereas, the velocity in the x-direction is constant: $\frac{dx}{dt} = v_{0x}$.

        • The position of the object in y-direction is a parabola. The position of the object in x-direction is a linear function.

     3. Test the solution

Solve Simple Harmonic

Problems:

1. Solve the equation of motion for simple harmonic motion

2. Oscillator Equation of Motion

Notes:

(2025-02-08)

1. Two properties of the point of stable equilibrium:

   1. Net force is 0

   2. The force is pulling back

2. Example of a spring connected to wall and a mass

   • Physical situation and free body diagram:

   • Apply the Two Steps:

     1. Apply Newton’s second law to write Equation Of Motion:

        $$ -k \cdot x(t) = ma $$

        • Hooke’s law: The force exerted by a spring is equal to the spring constant $k$ (in $N/m$) multiplied by the displacement from its natural length. The negative sign indicates that the force acts in the direction opposite to the displacement.

        • Use some notations for convenience:

          $$ \begin{aligned} x &= x(t) & \text{A function of time} \\\ \dot{x} &= \frac{dx}{dt} & \text{First derivative} \\\ \ddot{x} &= \frac{d^2 x}{dt^2} & \text{Second derivative} \end{aligned} $$

          Therefore, the EOM can be written as:

          $$-kx = m \ddot{x}$$

     2. Solve the differential equation to find out $x$

        • Cannot integrate this equation w.r.t. time becuase the exact form of $x$ is unknown.

          (2025-03-13)

          • Comparision:

            Situation	Constant force	Varying force
            EOM	-mg=ma	-kx=ma
            Feature	Force’s 原函数是一次函数	Force varies along with x
            Solving x	Integrate directly w.r.t. $t$	Cannot integrate as $x(t)$’s form is unknown

        1. Guess $x(t)$

           Which functions satisfy the condition that their second derivatives are themselves ($\ddot x = x$)?

           1. $x = \boxed{ A sin(Bt + C) }$.
              • $\dot{x} = B Acos(Bt + C)$,   $\ddot{x} = -B^2 \boxed{ Asin(Bt + C) }$ (using chain rule)
           2. $x = \boxed{ A cos(Bt + C) }$,
              • $\dot{x} = -B Asin(Bt + C)$,   $\ddot{x} = -B^2 \boxed{ Acos(Bt + C) }$
           3. $x = \boxed{ Ae^{(Bt + C)} }$ (Exponential function)
              • $\dot{x} = B Ae^{(Bt + C)} $,   $\ddot{x} = B^2 \boxed{ Ae^{(Bt + C)} }$,
           4. $x = 0$ (Constant function)

           • I tried to look for potential functions among “elementary functions”, however, I found they are not able to cover all functions in the world.

           • Since $cos(t) = sin(t+π/2)$, the 2nd derivative of $cos(t)$ equals the 2nd derivative of $sin(t+π/2)$.

           (2025-03-18)

           • 三角函数: sinx, cosx 的二阶导等于它的相反数， 指数函数的二阶导等于它 自身：$\frac{d^2(e^x)}{dx^2}=e^x$

           • If x(t) = sin(t), cos(t), or $e^t$ and plug the x(t) in the EOM: $-kx(t) = m \ddot{x(t)}$, the x(t) and $\ddot{x(t)}$ will be cancelled, leaving the relationship between k and m.

           • For sin(t) and cos(t), the spring constant k differs from m by a factor of B^2. For e^t, k is -B^2 times m.

             B is the angular frequency = $\sqrt{k/m}$ or $\sqrt{-k/m}$.

        2. Substitute a possible solution: $x = Asin(Bt + C)$ function into the EOM to find conditions that make the guessed $x$ one of possible solutions.

           $$ \begin{aligned} -k Asin(Bt + C) &= m \cdot [ -AB^2 sin(Bt + C) ] \\ -k \cancel{ Asin(Bt + C) } &= m \cdot [ \cancel{-A} B^2 \cancel{ sin(Bt + C) } ] \\ k &= m B^2 \end{aligned} $$
           • As long as $B$ is a specific value: $\sqrt{\frac{k}{m}}$, while A and C can be any values, the $x$ is equal to $A sin( \sqrt{\frac{k}{m}}t + C)$

           (2025-03-19)

           • The B in $x(t) = Asin(Bt+C)$ needs to be $\sqrt{\frac{k}{m}}$ to make the EOM: $-kx = ma$ satisfied.

             $$ k = B^2 m \\\ k = (\sqrt{\frac{k}{m}})^2 m $$

3. Mnemonics

   • Actions:

     (2025-07-20T13:05)

     1. 因为 x"(t) 的具体形式未知，所以无法通过对 x"(t) 直接积分以求出 x(t)

     2. 解为 A sin(Bt+C) 或 Acos(Bt+C) 时，B 是“背头金”，B = $\sqrt{k/m}$

     3. 三角函数的二阶导等于它的相反数乘以系数 B^2，而指数函数的二阶导等于它自己乘以系数 B^2

        如果 x(t)=Asin(Bt+C) 或 Acos(Bt+C)，则 x"(t)= -B^2 x(t)，所以 -kx(t) = mx"(t) -> k= mB^2

        如果 x(t)=A e^{Bx+C}，则 x"(t) = B^2 x(t)，所以 -kx(t) = mx"(t) -> -k = mB^2

     4. 三角函数可以和 e 指数函数通过欧拉公式互相转化：e^{ix} = cosx + isinx， 但是其中涉及到 i，这是两种解的 B 相差一个 i 的原因吗？

Quiz:

Is $Asin(Bt+C)+Dcos(Bt+C)$ a possible solution to the simple harmonic oscillator equation of motion?

Answer: Yes

• I think there are two criteria need to be satisfied for a possible solution:

  1. Its secondary derivative is equal to itself multiplied by some factors, becuase this way the $x$ itself can be cancelled.

  2. The conditions that derived from plugging the guessed x into EOM can be met.

• First, the proposed function’s second derivative could be equal to the scaled itself.

  1. The first derivative:

     $\dot{x} = AB cos(Bt + C) - DB sin(Bt + C)$

  2. The second derivative:

     $$ \begin{aligned} \ddot{x} &= -AB^2 sin(Bt + C) - DB^2 cos(Bt + C) \\ & = -B^2 [A sin(Bt+C) + D cos(Bt+C)] \\ & = -B^2 x \end{aligned} $$
  No need to be exactly the same as x

  • The $x$ cannot become $\ddot{x}$ by performing add or multiplication operations.

    The $x$ can be $-\ddot{x}$ if B=1.

    $$ \begin{aligned} x &= \boxed{ Asin(t+C)+Dcos(t+C) } \\ \ddot{x} &= - ( \boxed{ Asin(t+C)+Dcos(t+C) } ) = -x \end{aligned} $$

  So the first criterion is satisfied.

• Second, plugging the $x(t)$ in the EOM for simple harmonic motion:

  Old notes

  $$ \begin{aligned} -kx &= m \ddot{x} \\\ -k [Asin(Bt+C) + Dcos(Bt+C)] &= -AB^2 m sin(Bt + C) - DB^2 m cos(Bt + C) \\\ (AB^2m-kA) sin(Bt+C) &= (kD-DB^2m) cos(Bt+C) \\\ \end{aligned} $$

  There are two conditions to be satisfied to make the above equation held:

  1. Same amplitude:

     $$ AB^2m-kA = kD-DB^2m \\\ B = \sqrt{\frac{k}{m}} $$

  2. Phase requirement: When $θ = \frac{π}{4} + nπ$, sin(θ) = cos(θ)

  • 有没有可能是因为 cos(θ) = sin(θ+pi/2)，就可以把 cos(θ) 合进 sin 里面呢？那样就变成了一个 sin 函数
  $$ \begin{aligned} -kx = m \ddot x \\ -kx = -mB^2x \\ k = B^2m \\ B = \sqrt{\frac{k}{m}} \end{aligned} $$

Oscillator Solution

1. Four Parameters: A,B,C,T

   • References:

     1. Module 1 > Oscillator Solution | Coursera

   ---

   • Supports:

     1. A is Amplitude

     2. C is Phase lag: The distance between the first t’ where sin(t’)=0 and the zero tick on the time axis

        To write the EOM on each bead, think of each bead at the zero tick of the time axis.

        • 取决于时间轴零点的选取。相位滞后，但其实是坐标轴（零点）向前移动了 phi

        • Phase lag is caused by the difference in the selection of the origin.

        (2025-07-23T07:21)

        • C 决定了 时间轴零点 的选取 C 由时间零点决定，C 是相对于时间零点测量的。

          C = pi/3, 就是说波形的起点在所选取的时间零点的左边 pi/3 处。

          “Phase lag has to do with where the sinusoid sits relative to the time origin.”

        • 相位滞后本质上是时间的滞后：波上的同一点需要慢 C 秒才能传到零点。

          一个经过调制的波相对于无相位偏移的正弦波慢了 C 秒，

          “Phase lag is the offset in the part with a time in it.”

        (2025-07-24T23:47)

        • 观测到的 一段 波形与波形的起点相关，波形的 表达式（phase lag）取决于时间轴零点的选取。

          画图 3 个波：sin(t), sin(t+pi/3), sin(t+2pi/3)

     3. T = 2pi/B

     4. B is angular frequency

     • $\sqrt{k}{m}$ is natural frequency

     5. Plot of sinusoid: $Asin(Bt+C)$
   mindmap Oscillator Frequency Amplitude, Phase Lag

   (2025-06-17T00:47)

   Does “phase” mean trend? no only the value at a point, but the rising and falling trend also are a factor of phase. And also the rising/falling speed included?

   • sin(-pi/3) is the original sin(0)

   • cos(pi/3) is the original cos(0)

   • sin(t) = cos(t+pi/2)

2. B is independent on A and C

   • Supports:

     1. Frequency is independent to amplitude and phase lag, but determined by spring constant and mass.

summary

1. Kinematics -> Constant force -> Harmonic -> Sin Solution

   • Supports:

     (2025-08-06T07:13)

     1. Important conclusions