A string of length \( L \) fixed at both ends has, for its fundamental mode, a wavelength of
Answer: A
Fixed ends force nodes at both, and the fundamental has no interior node, so the length holds half a wave:
\[ L \;=\; \frac{\lambda}{2} \quad \Longrightarrow \quad \lambda \;=\; 2L. \]
The frequency is then \( c/2L \). A quarter wave in the length would need one free end, as in a closed organ pipe
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A string of length 2 m carries waves at 40 m/s. Its fundamental frequency is
Answer: D
The fundamental wavelength is twice the length, so \( \lambda = 4 \) m, and
\[ f \;=\; \frac{c}{\lambda} \;=\; \frac{40}{4} \;=\; 10\ \mathrm{Hz}. \]
Equivalently \( f = c/2L = 40/(2 \times 2) \). Taking the wavelength as the length itself rather than twice it would double the answer to 20 Hz
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In the heat equation \( \partial u/\partial t = \alpha\,\partial^{2}u/\partial x^{2} \), the constant \( \alpha \) has dimensions of
Answer: C
For the two sides to agree dimensionally, \( \alpha \) must carry the dimensions of the left side divided by those of \( u/x^{2} \), that is
\[ \left[\alpha\right] \;=\; \frac{\Theta/T}{\Theta/L^{2}} \;=\; \frac{L^{2}}{T}, \]
units of square metres per second. This is why a diffusion length grows as the square root of time rather than in proportion to it
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Three of these four describe the one-dimensional heat equation. The exception is
Answer: B
Finite-speed propagation along characteristics belongs to the hyperbolic wave equation. The classical heat equation is parabolic, its solutions smooth with time, and it does require both an initial distribution and conditions at the ends. Its disturbances spread instantly in principle, which is a known shortcoming of the model
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Match the equation with its type. 1. \( z_{xx} + z_{yy} = 0 \) 2. \( z_{t} = \alpha z_{xx} \) 3. \( z_{tt} = c^{2}z_{xx} \) P. Parabolic Q. Elliptic R. Hyperbolic
Answer: A
The discriminant \( B^{2} - 4AC \) is \( -4 \), zero and \( 4c^{2} \) respectively. Negative gives elliptic, with data round a closed boundary; zero gives parabolic, marching forward from an initial state; positive gives hyperbolic, carrying disturbances along characteristics at finite speed
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Match the first-order equation with the method it calls for. 1. \( Pp + Qq = R \) 2. \( z = px + qy + f(p,q) \) 3. Neither of the standard forms P. Replace \( p \) and \( q \) by constants Q. Subsidiary equations \( dx/P = dy/Q = dz/R \) R. Charpit's auxiliary equation
Answer: D
Lagrange's linear form is integrated through its subsidiary equations. Clairaut's form needs nothing more than replacing the two derivatives by constants. Anything outside the standard types goes to Charpit's method, which is general but longer
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Put these steps of solving Lagrange's linear equation in order. 1. Write the general solution as an arbitrary function relating the two integrals 2. Identify \( P \), \( Q \) and \( R \) from the equation 3. Find two independent integrals of the subsidiary equations 4. Write the subsidiary equations
Answer: C
The three coefficients are read off first, then written as the subsidiary equations. Two independent integrals \( u = a \) and \( v = b \) follow, and the general solution is \( \phi(u,v) = 0 \) for an arbitrary \( \phi \). Writing the solution before the integrals are found leaves nothing to put inside the function
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Consider the following statements about partial differential equations. 1. Eliminating an arbitrary function raises the order by one for each function 2. The general solution contains arbitrary functions rather than only constants 3. The type of a second-order equation follows from the sign of \( B^{2} - 4AC \)
Answer: B
All three hold. The presence of arbitrary functions is the essential difference from the ordinary case, and it is why one boundary curve rather than a few numbers is needed to pin a solution down. The discriminant test mirrors the classification of conics, and for the same algebraic reason
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Consider the following statements about the wave equation. 1. Its general solution is a sum of two travelling waves 2. It is hyperbolic 3. A single travelling wave can satisfy both initial conditions
Answer: A
The first two hold. The third fails: two initial conditions, the initial displacement and the initial velocity, need two arbitrary functions to satisfy, so one travelling wave is not enough unless the two conditions happen to be compatible with it. That is exactly why d'Alembert's solution carries both terms
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Just one of these four is a linear partial differential equation. Which?
Answer: D
Linearity requires the unknown and its derivatives to appear only to the first power and never multiplied by one another; coefficients depending on \( x \) and \( y \) are unrestricted. The squares and the products \( pq \) each break that, and it is those cases that need the nonlinear machinery of the standard types and Charpit's method
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