Question

A string of length 1 meter is fixed at both ends and obeys the wave equation \(u_{tt}=4u_{xx}\) with initial conditions: \(u(x,0)=\sin(\pi x)\), \(u_t(x,0)=0\). Then its solution \(u(x,t)\) is ____.

• A. \(\sin(\pi t)\cos(\pi x)\)
• B. \(\sin(2\pi t)\cos(\pi x)\)
• C. \(\sin(2\pi x)\cos(\pi t)\)
• D. \(\sin(\pi x)\cos(2\pi t)\)

Hint:

Separate variables; the initial shape \(\sin(\pi x)\) is exactly the first normal mode, so only n=1 survives with c=2.

Answer 1

The Correct Option is D

Step 1: Since \(u_t(x,0) = 0\), d'Alembert's solution reduces to

\[u(x,t) = \frac{1}{2}\left[F(x+ct) + F(x-ct)\right]\]

where \(F\) is the odd, period 2 extension of the initial shape \(u(x,0) = \sin(\pi x)\), and \(c = 2\).

Step 2: The function \(\sin(\pi x)\) is already odd and has period 2 (since \(\sin(\pi(x+2)) = \sin(\pi x)\)), so it is its own extension: \(F(s) = \sin(\pi s)\).

Step 3: Substitute:

\[u(x,t) = \frac{1}{2}\left[\sin(\pi(x+ct)) + \sin(\pi(x-ct))\right]\]

Step 4: Use the identity \(\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\) with \(A = \pi(x+ct)\), \(B = \pi(x-ct)\), so \(\frac{A+B}{2} = \pi x\) and \(\frac{A-B}{2} = \pi c t\):

\[u(x,t) = \frac{1}{2}\cdot 2\sin(\pi x)\cos(\pi c t) = \sin(\pi x)\cos(\pi c t)\]

Step 5: Substitute \(c = 2\):

\[u(x,t) = \sin(\pi x)\cos(2\pi t)\] \[\boxed{u(x,t) = \sin(\pi x)\cos(2\pi t)}\]