Hooke's Law & Simple Harmonic Mechanical Oscillation

Source

MIT Professor Lewin

Springs

Hooke's Law

• If we displace the spring a distance of x, then $$F_{spring-restoring} = -k x $$
• This is also known as Hooke's Law.

Measuring the spring constant k

• $$k = \frac{\Delta F}{\Delta X} $$

• $$T_{period-of-oscillation} = 2\pi\sqrt\frac{m}{k} $$
• Knowing the mass and the period T, $$ k = m * \left(\frac{2\pi}{T}\right)^2 $$
• Also knowing period T and spring constant k, $$ m = k * \left(\frac{T}{2\pi}\right)^2 $$
• Notice these measurements do not depend on gravity and can be done in  the weightlessness of space.
• $$ m a =  -k x $$
• $$ m \ddot{x} = -k x$$
• $$ m \ddot{x} + k x = 0$$
• $$ \ddot{x} + x \frac{k}{m}  = 0$$
• Our test function for x is:
• $$ x = A \cos (\omega t + \phi)$$
• A is the amplitude of the oscillation
• omega is the angular frequency in radians/sec
• psi is the phase angle in radians 
• If the time t is advanced to T, then omega * T is 2 pi = 360 degrees = one full cycle or period
• $$ T = \frac{2 \pi}{\omega} $$
• $$ f = \frac{1}{T} ~ \mbox{hertz}$$
• $$ \dot{x} = - A \omega \sin (\omega t + \phi)$$
• Derivative is with regard t
• Because the derivative of the cos is the -sin 
• And because omega is multiplied time t
• $$ \ddot{x} = - A \omega^2  \cos (\omega t + \phi)$$
• The derivative of the sin is the cos 
• $$ \ddot{x} = - A \omega^2  \cos (\omega t + \phi)$$
• Already started with $$ x = A \cos (\omega t + \phi) $$
• $$ \ddot{x} = - \omega^2  x$$
• We already have: $$ \ddot{x} + x \frac{k}{m}  = 0$$
• $$ - \omega^2  x + x \frac{k}{m}  = 0$$
• $$ \omega^2 =  \frac{k}{m}  = 0$$
• $$\omega = \sqrt \frac{k}{m} $$
• $$T = 2 \pi \sqrt\frac{m}{k} $$

Pendulum

• l = length of the string
• theta = the angle the pendulum is displaced from the vertical

• In the x direction: $$m a = m \ddot{x} = - T_\theta \sin\theta = - T_\theta \frac{x}{l} $$
• In  the y direction: $$ m \ddot{y} = T_\theta \cos\theta - m g $$
• These two equations are unsolvable as they are. But now we will make some approximations.
• We will use "Small Angle Approximations"
• Theta must remain a small angle. So theta must be << 1.
• Cosine of theta must be very close to 1; and sine theta will be very close to 0.
• 5 degrees = 0.996
• 10 degrees = 0.985
• The y acceleration is approximately zero
• The y excursion at 5 degrees is about 4% of x direction
• The y excursion at 10 degrees is about 9 % of x direction
• Revisiting our previous unsolvable equations
• $$ m \ddot{y} = T_\theta \cos\theta - m g $$
• Since cosine of theta, $$ 0 = T -mg $$
• $$ T = m g $$
• $$ m \ddot{x} = - T_\theta \sin\theta = - T_\theta \frac{x}{l} $$
• $$ m \ddot{x} + m g \frac{x}{l} = 0 $$
• $$ \ddot{x} + \frac{g}{l} x = 0 $$
• Notice the similarity of the spring equation and the pendulum equation
•  $$ \ddot{x} + \frac{k}{m} x  = 0  \mbox{for a spring}$$
•  $$ \ddot{x} + \frac{g}{l} x = 0  \mbox{for a pendulum}$$
• Since we have solved  the equation for the spring, the equations by extension for the pendulum are:
• $$ x = A \cos (\omega t + \phi)$$ 
• $$ omega = \sqrt \frac{g}{l} $$
• $$ T = 2 \pi \sqrt \frac{l}{g} $$