Chapter 10: Problem 18
Suppose that the end \(x=0\) of a uniform bar with crosssectional area \(A\) and Young's modulus \(E\) is fixed, while the longitudinal force \(F(t)=F_{0} \sin \omega t\) acts on its end \(x=L\), so that \(A E u_{x}(L, t)=F_{0} \sin \omega t\). Derive the steady periodic solution $$ u(x, t)=\frac{F_{0} a \sin (\omega x / a) \sin \omega t}{A E \omega \cos (\omega L / a)} . $$
Short Answer
Step by step solution
Understand the Problem
Establish the Wave Equation
Apply Boundary Conditions
Assumption for Steady-State Solution
Differentiate and Substitute
Solve the Spatial Equation
Apply Initial Conditions to Spatial Component
Use Free-End Boundary Condition
Solve for Coefficient A
Assemble the Final Solution
Verify the Solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Wave Equation
Steady Periodic Solution
Boundary Conditions
- The fixed end condition, \( u(0, t) = 0 \), implies that there is no displacement at one end of the bar, ensuring it is anchored tightly.
- At the free end \( x = L \), the force condition \( A E \frac{\partial u}{\partial x}(L, t) = F_0 \sin(\omega t) \) ties the displacement gradient to the applied periodic force, conveying how force alters the bar's displacement at that point.