Solving for Simple Harmonic Motion: A Picture Problem

In summary, the conversation discusses the equation for simple harmonic motion and how to determine its period. The equation is of the form ##~\ddot y=-\omega^2 y~## and any equation in that form describes simple harmonic motion. By plugging in the most general solution, the period of the motion can be deduced.
  • #1
zenterix
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Homework Statement
An electric dipole lying in the xy-plane with a uniform electric field applied in the positive x-direction is displaced by a small angle ##\theta## from its equilibrium position. The charges are separated by a distance ##2a##. The moment of inertia of the dipole about the center of mass is ##I_{cm}##.

If the dipole is released from this position, show that its angular orientation exhibits simple harmonic motion. What is the period of oscillation?
Relevant Equations
$$\vec{\tau}=-2aE_x\sin{\theta} \hat{k}$$

$$\tau_z=I_{cm}\alpha_z$$

$$\alpha_z=\frac{\tau_z}{I_{cm}}=\frac{-2aE_x\sin{\theta}}{I_{cm}}=-B\sin{\theta}$$

where $$B=\frac{2aE_x}{I_{cm}}$$

This is a differential equation. I think that solving this equation would provide the correct result, but I don't want to go this route.

One route is, if ##\theta## is small, to use the approximation ##\sin{\theta}\approx\theta##.

Then

$$\alpha_z(t)=\theta''(t)=-B\theta(t)$$

I think this is the differential equation representing a simple harmonic motion of an ideal spring with

$$B=\frac{k}{m}$$
Here is a picture of the problem

1674603667180.jpeg


It is not clear to me how to really prove that the equation for ##\theta(t)## is simple harmonic motion, and what the period of this motion is.
 
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  • #2
Your equation is the equation for simple harmonic motion which is of the form ##~\ddot y=-\omega^2 y~## where ##\omega## is a real constant and the differentiation is with respect to time. Any equation in that form describes simple harmonic motion. Do you know what the most general solution of that is? If so, plug it in and you should be able to deduce the period of the motion.
 

FAQ: Solving for Simple Harmonic Motion: A Picture Problem

What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. It is characterized by oscillatory motion, such as that of a mass on a spring or a pendulum.

How do you derive the equation of motion for SHM?

The equation of motion for SHM can be derived from Newton's second law. For a mass m attached to a spring with spring constant k, the restoring force is F = -kx. According to Newton's second law, F = ma, where a is the acceleration. Hence, ma = -kx, or a = -(k/m)x. Since acceleration a is the second derivative of displacement x with respect to time t, we get the differential equation: d²x/dt² + (k/m)x = 0. This can be solved to give x(t) = A cos(ωt + φ), where ω = √(k/m) is the angular frequency, A is the amplitude, and φ is the phase constant.

What are the key parameters in SHM?

The key parameters in SHM are the amplitude (A), which is the maximum displacement from the equilibrium position; the angular frequency (ω), which determines how fast the oscillations occur; the phase constant (φ), which determines the initial position and velocity of the oscillating object; and the period (T), which is the time taken for one complete cycle of oscillation.

How do you calculate the period and frequency of SHM?

The period (T) is the time it takes for one complete cycle of oscillation. It is given by T = 2π/ω, where ω is the angular frequency. The frequency (f) is the number of cycles per unit time and is the reciprocal of the period, f = 1/T. For a mass-spring system, ω = √(k/m), so the period T = 2π√(m/k) and the frequency f = 1/(2π√(m/k)).

What is the energy in SHM and how is it conserved?

In SHM, the total mechanical energy is conserved and is the sum of kinetic energy (KE) and potential energy (PE). The kinetic energy is given by KE = 1/2 mv², where v is the velocity, and the potential energy in a spring system is PE = 1/2 kx², where x is the displacement. At maximum displacement (amplitude), the energy is all potential, and at the equilibrium position, the energy is all kinetic. The total energy E = 1/2 kA² remains constant throughout the

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