Mechanical Frequencies Increase in Small Systems

Mechanical resonance frequencies depend on the dimensions of the system at hand. For the simple pendulum, o) = (g/l)1/2, where I is the length of the pendulum rod and g is the acceleration of gravity. The period T = 2jt/&> of the pendulum of a grandfather clock is exactly one second, and corresponds to a length I about one meter (depending on the exact local value of the gravitational acceleration, g, which is approximately 9.8 m/s2). The relation co= (g/l)1/2 indicates that the period of the pendulum scales as V/, so that for I one micron T is one ms; a one-micron sized grandfather clock (oscillator) would generate a 1000 Hz tone. Either clock could be used (in conjunction with a separate frequency measurement device or counter), to measure the local vertical acceleration a = d2y/dt2 according to a> = [(g+fl)/i]1^2-

If used in this way as an accelerometer, note that the miniature version has a much faster response time, 1ms, than the original grandfather clock, which would have a time resolution of about a second.

A mass m attached to a rigid support by a spring of constant k has a resonance frequency (fc/m)1/2. This oscillator, and the pendulum of the grandfather clock, are examples of simple harmonic oscillation (SHO).

Simple harmonic oscillation occurs when a displacement of a mass m in a given direction, x, produces an (oppositely directed) force F - -he, The effective spring constant k has units of N/m in SI units. According to Newton's Second Law (1.2) F = ma « md2x/dt2. Applied to the mass on the spring, this gives the differential is a solution of the equation for arbitrary amplitude xinax and arbitrary initial phase angle Ô, but only when d2x/dt2+(k/m)x=0. x = xmiLX cos(wf + f>)

Hie period of the motion is therefore T **2%(m/k)lf2. The maximum values of the speed v « dxfdt and the acceleration d2x/dt2 are seen to be xmSiXw and Xm^cd2, respectively. The total energy E = t/ + K in the motion is constant and equal to (In nanophysics, which is needed when the mass m is on an atomic scale, the same frequency = (k/m)112 is found, but the energies are restricted to En~ (n + 1/2) Urn, where the quantum number n can take zero or positive integer values, fi = kjln, and Planck's constant is h = 6.67 x 10"34foule • s.)

Simple harmonic oscillation is a more widely useful concept than one might think at first, because it is applicable to any system near a minimum, say xQJ in the system potential energy U(x). Near the potential energy U{x) can be closely approximated as a constant plus h(x-x0)2/2, leading to the same resonance frequency for oscillations of amplitude A in x-x0. An important example is in molecular bonding, where is the interatomic spacing.

More generally, the behavior applies whenever the differential equation appears, and the resonant frequency will be the square root of the coefficient of x in the equation. In the case of the pendulum, if xzL is the horizontal displacement of the mass m., then F x/L and w = (g/L)1/2,

Considering the mass and spring to be three-dimensional, mass m will vary as I3 and k will vary as I, leading to oj^ a IT1. Frequency inversely proportional to length scale is typical of mechanical oscillators such as a violin or piano string and the frequency generated by a solid rod of length L struck on the end. In these cases the period of the oscillation T is the time for the wave to travel 21, hence T = 2L/v. (This is the same as L = A/2, where X=vT is the wavelength. If the boundary conditions are different at the two ends, as in a clarinet, then the condition will be L= A/4 with half the frequency.) Hence co = 2k{u/L) where v=(F/p)1/2 for the stretched string, where F is the tension and p the mass per unit length.

The speed of sound in a solid material is v = (Y/p)1/2, with Ythe Young's modulus. Young's modulus represents force per unit area (pressure stress) per fractional deformation (strain). Young's modulus is therefore a fundamental rigidity parameter of a solid, related to the bonding of its atoms. For brass, Y = 90GPa = 90xl09N/m2. (This means, e.g., that a pressure F/A of 101 kPa applied to one end of a brass bar of length I = 0.1m would compress its length by Al = LF/YA = 11 |im.) Note that Y = 90GPa and p=104kg/m3, values similar to brass, correspond to a speed of sound v = 3000m/s. On this basis the longitudinal resonant frequency of a 0.1m brass rod is/ =v/2L = 15 kHz. This frequency is in the ultrasonic range.

If one could shorten a brass rod to 0.1 micron in length, the corresponding frequency would be 15 GHz, which corresponds to an electromagnetic wave with 2 cm wavelength. This huge change in frequency will allow completely different applications to be addressed, achieved simply by changing the size of the device!

A connection between macroscopic and nanometer scale descriptions can be made by considering a linear chain of N masses m spaced by springs of constants K, of length a. The total length of the linear chain is thus L = Na.

0 0

Post a comment