Question

Atoms in a solid are not stationary, but vibrate about their equilibrium positions. Typically, the frequency of vibration is about f = 4.60 x 1012 Hz, and the amplitude is about 3.55 x 10-11 m. For a typical atom, what is its (a) maximum speed and (b) maximum acceleration?

Answer 1

We can find the maximum speed and acceleration by considering the general formula for simple harmonic motion.

Simple harmonic motion is defined as the motion of a particle experiencing an acceleration proportional to its distance from a point, with the acceleration pointing towards the point.

In essence: a = - k x

A solution to this second order differential equation is through the usage of trigonometrical identities, and we will exploit the unique characteristics of these trigonometrical identities to calculate the maximum speed and acceleration of the particle.

Firstly, let us examine what we already have. The frequency of the particle has been established at 4.60x10^12 hertz. Now using this frequency, we can introduce a useful parameter, the angular frequency. The angular frequency will be used in the calculations later.

We define the angular frequency to be 2π multiplied by the frequency.

w = 2 π f

Now a possible solution to the equation a = - k x is:

x = A sin (wt)

Where x is the position of the particle at time t, A is the amplitude of the particle, and w is the angular frequency. We can replace the w in the equation with 2πf, to obtain:

x = A sin ( 2πf t )

Now let us examine this equation closely. There is a sine function in the equation. Now we know that the sine of any number can only take a value between -1 and 1. This means that sin (2πf t) can only be between -1 and 1. When we consider this interesting observation in the equation, we notice that x can only take values between -A and A, which is consistent with the definition of an amplitude.

"A" represents the maximum displacement of the particle.

Now we can apply a similar argument to the speed and acceleration of the particle.

We know that velocity v=dx/dt, the first time differential of the displacement.

So, let us differentiate our equation with respect to time.

x= A sin (2πf t)
v = dx/dt = (2πf) A cos (2πf t)

Notice that we have taken out the 2πf inside the sine function as part of the chain rule, and the sine has been differentiated to cosine.

Upon inspection, we notice something similar to the maximum displacement argument. cos ( 2πf t) can only take values between -1 and 1. Also, 2πf and A are both constants.

Thus, the maximum velocity occurs when cos ( 2πf t) = 1

v(max) = ( 2πf ) A (1)

With f=4.6x10^12, and A = 3.55x10^-11 m, we obtain v(max) = 1026.04 ms^-1

Now a=dv/dt. Differentiating the velocity equation one more time:

v = ( 2πf ) A cos ( 2πf t)
dv/dt = - ( 2πf )² A sin ( 2πf t)

Now we have a negative sign in front. This is expected, because of the definition of simple harmonic motion: a= -kx. Remember that x= A sin ( 2πf t), so the acceleration is simply a = - ( 2πf )² x

But that aside, we also notice that sin (2πf t) is bound between -1 and 1 again. This time, the maximum acceleration occurs when sin ( 2πf t) = -1. Thus,

a(max) = ( 2πf )² A

Subbing in the values, we obtain a(max) = 2.96 x 10^16 m/s^2