Classical Vibrating String

Contents


Problem Setup

Consider a flexible string held stationary at both ends and free to vibrate transversely subject only to the restoring forces due to tension in the string. Deduce the DE for such systems and define all parameters that distinguish the systems. Do this with a string of normal length l and using the coordinates depicted here.

Figure showing coordinates and defining symbols for the transverse vibrating string. vibr_string.gif (1881 bytes) The length of the string is l, distance along the string from the left end is x, its displacement or wave form is y(x,t).

The Differential Equation

Our hypothesis is that the string is under constant tension, T, established when it was stretched between its fixed end points. The transverse displacement at position x along the string at time t is denoted y(x,t). Our aim is to deduce the partial differential equation (PDE) governing y(x,t).

The tangent to the string at y(x,t) is indicated in the figure; it forms the hypotenuse of a right triangle with legs parallel and perpendicular to the undisplaced string position. Also shown is a copy of the tangent triangle (legs dx, dy, and hypotenuse ds) and a similar triangle with legs Tx (the component of tension along the x-direction), Ty (that parallel to displacement) and hypotenuse T (tension in the string). If we suppose that displacements are small then Tx is approximately equal to T; but Ty is given by Ty = T y/x.

Consider a short segment of the string, length dx, at x where its displacement is y(x,t). The mass of such a segment is mdx where m is the linear mass density of the string. Denote the acceleration of this segment as y"(x,t) = 2y/t2. According to Newton's laws, the acceleration is determined by the net force on the segment: m y"(x,t) = F. Now the net force is the difference in Ty at the left and right ends of the segment dx; namely, F = dTy = d/dx (T y/x) dx = T 2y/x2 dx. Then the DE for the vibrating string becomes m2y/t2 = T 2y/x2. Only the ratio T/m is important to the problem and it is denoted by c2 = T/m. Since c has units length over time we expect this to be the wave velocity. Finally, we are to solve

2y/t2 = c2 2y/x2,

the one-dimensional wave equation (a partial differential equation, second order in each of t and x). The boundary conditions are y(0) = y(l) = 0, and initial conditions are y(x,0) and y'(x,0).

The General Solution

The vibrating string problem can be solved using the method of separation of variables. Since y is a function of x and t, we look for a solution in the form of a product, y(x,t) = X(x)T(t). Upon substitution into the wave equation, the product form requires that X(d2T/dt2) = c2 T(d2X/dx2). Now the variables can be separated if both sides are divided by the Product X(x)T(t): wpe2C.gif (1270 bytes). The left hand side, a function of t alone, is equal to the right hand side, a function of x alone, for all values of x and t. For the independent variables x and t, this is only possible if both sides are equal to the same constant, say -f2. Then the original 2nd order PDE becomes two 2nd order ODEs coupled through the separation constant f: wpe2E.gif (1520 bytes).

Each of these ODEs has general solutions in circular functions: X(x) = Asin(f x/c) + Bcos(f x/c) and T(t) = Csin(ft) + Dcos(ft), where A, B, C, and D are arbitrary integration constants.

The product form of solution is called a "standing wave" for the following reasons. For given f, A, and B, the factor X(x) is a wave shape giving the string displacement at x. For the same f and for given C and D, the factor T(t) oscillates with angular frequency f alternating between positive and negative values. Thus, the wave shape, X(x) does not change with time, only its amplitude oscillates as dictated by T(t).

Satisfying the Initial Conditions

Fundamental and overtones or harmonics

With both ends of the string tied to rigid walls, the displacements at the boundaries are given by y(0,t) = y(l,t) = 0. From the left boundary we are led to B(Csin(ft)+Dcos(ft)) = 0. Supposing that the T(t)0, it means that B=0. From the right boundary condition we are led to Asin(fl/c)[Csin(ft)+Dcos(ft)] = 0. Again, the time factor is not zero so that the frequency has to satisfy sin(fl/c) = 0 or f = npc/l where n is an integer. Standing wave solutions, those that factor according to X(x)T(t), exist only for "natural", normal, characteristic or eigen- frequencies determined by the string length and wave speed (l, and c). When n=1 the standing wave is the "fundamental", when n=2 it is the first overtone, and so forth.

General Solution

In general, the displacement of the vibrating string is given by
y(x,t) = wpe2F.gif (2056 bytes),
a superposition (linear combination) of many natural modes of vibration of the string. Since the sine and cosine functions are independent, it is always possible to determine the values of Cn and Dn from the initial wave form y(x,0), and initial wave velocity y'(x,0) using

wpe32.gif (1869 bytes), respectively. Clearly, this amounts to two applications of the Fourier series to find the Fourier coefficients Dn and Cn.


Examples

  1. Plucked string. A uniform string under tension and of length L is displaced at its center a distance h and released from initial velocity zero. Find its displacements as a function of time y(x,t). That is, the initial velocity is y'(x,0) = 0 while its initial displacement is y(x,0) = 2hx/L for 0<x<L/2 and y(x,0) = 2h(L - x)/L for L/2<x<L. This is a model for plucked strings on a harp or guitar. Answer: y(x,t) = wpe35.gif (2512 bytes). Note that the odd overtones are present but even overtones are absent; the amplitudes alternate in sign and decrease with the square of the overtone frequency.
  2. Struck string. In contrast to a plucked string, the struck string is initially not displaced but has initial velocity imparted by a hammer applied at some segment. Suppose a hammer having width s strikes the string at a position d so that the initial string velocity is y'(x,0) = wpe37.gif (1621 bytes) starting from resting position y(x,0) = 0. Find its displacements as a function of time y(x,t). Answer: y(x,t) = wpe36.gif (1927 bytes). In this case note that both even and odd overtones are present; the amplitudes are all of the same sign and decrease with the square of the frequency.

Problems

  1. A uniform steel piano string of length 5 feet is under a tension of 900 pounds throughout its length. The wire has linear density 0.027 lb/ft and cross sectional radius of 0.05 in. (a) Calculate the velocity of transverse waves in the string, c. (b) What is the fundamental frequency of vibration of this string?
  2. A uniform string with length L under tension is plucked at x = L/3 with an amplitude h and released. Find the resulting motion y(x,t).

References


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Created or up-dated 08/03/99   by R.D. Poshusta
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