answer:To solve this problem, we can use the method of separation of variables. We assume that the solution can be written as a product of two functions, one depending only on x and the other depending only on t: u(x,t) = X(x)T(t). Substituting this into the wave equation, we get: X(x)T''(t) = 9X''(x)T(t). Now, we divide both sides by X(x)T(t): frac{T''(t)}{T(t)} = 9frac{X''(x)}{X(x)}. Since the left side depends only on t and the right side depends only on x, both sides must be equal to a constant, which we will call -lambda^2: frac{T''(t)}{T(t)} = 9frac{X''(x)}{X(x)} = -lambda^2. Now we have two ordinary differential equations: 1. T''(t) = -lambda^2 T(t) 2. X''(x) = -frac{lambda^2}{9} X(x) The second equation has the general solution: X(x) = Asin(frac{lambda x}{3}) + Bcos(frac{lambda x}{3}). Applying the boundary conditions u(0,t) = u(L,t) = 0, we get X(0) = X(1) = 0. Since L = 1, we have: X(0) = Bcos(0) = B = 0, and X(1) = Asin(frac{lambda}{3}) = 0. Since A cannot be zero (otherwise the solution would be trivial), we must have frac{lambda}{3} = npi for some integer n. Thus, lambda = 3npi, and the solution for X(x) is: X(x) = Asin(3npi x). Now we solve the first equation for T(t). Since lambda = 3npi, we have: T''(t) = -(3npi)^2 T(t). The general solution is: T(t) = Ccos(3npi t) + Dsin(3npi t). Applying the initial condition frac{partial u}{partial t}(x,0) = 0, we get: frac{partial u}{partial t}(x,0) = X(x)T'(0) = 3npi X(x)D = 0. Since X(x) is not identically zero, we must have D = 0. Thus, the solution for T(t) is: T(t) = Ccos(3npi t). Now, the general solution for u(x,t) is a sum of solutions for each n: u(x,t) = sum_{n=1}^{infty} A_n sin(3npi x)cos(3npi t). To find the coefficients A_n, we use the initial condition u(x,0) = sin(pi x): sin(pi x) = sum_{n=1}^{infty} A_n sin(3npi x)cos(0). Using the orthogonality of sine functions, we can find A_n: A_n = 2int_0^1 sin(pi x)sin(3npi x) dx. For n = 1, we have: A_1 = 2int_0^1 sin(pi x)sin(3pi x) dx = frac{4}{pi}. For n > 1, the integral evaluates to zero. Therefore, the solution for the displacement of the string is: u(x,t) = frac{4}{pi}sin(3pi x)cos(3pi t).

answer:To find the wave equation that governs the displacement of the string, we need to know the linear mass density (μ) of the string. Since it is not given, we will assume it as a variable. The wave speed (v) on the string can be calculated using the formula: v = sqrt(T/μ) where T is the tension (16 N) and μ is the linear mass density. The wave equation for the displacement of the string (y) can be written as: ∂²y/∂t² = v² * ∂²y/∂x² Now, let's find the displacement of the string at a point 10 cm away from the end where the displacement is given, at a time of 0.01 seconds after the initial displacement. Since the string is initially at rest, we can assume that the displacement is caused by a wave pulse traveling along the string. The shape of the pulse is not given, so we cannot determine the exact displacement function. However, we can estimate the time it takes for the pulse to reach the point 10 cm away from the end. Using the wave speed formula, we have: t = x/v where x is the distance (10 cm = 0.1 m) and t is the time. We know that v = sqrt(16/μ), so we can rewrite the formula as: t = 0.1/sqrt(16/μ) Given that t = 0.01 seconds, we can solve for μ: 0.01 = 0.1/sqrt(16/μ) 0.01 * sqrt(16/μ) = 0.1 sqrt(16/μ) = 10 16/μ = 100 μ = 16/100 = 0.16 kg/m Now that we have the linear mass density, we can find the wave speed: v = sqrt(16/0.16) = sqrt(100) = 10 m/s However, without knowing the shape of the pulse or the boundary conditions, we cannot determine the exact displacement of the string at the given point and time.

answer:To solve the given wave equation with the given initial and boundary conditions, we can use the method of separation of variables. Let u(x, t) = X(x)T(t). Then, the wave equation becomes: X''(x)T(t) - (1/v^2)X(x)T''(t) = 0 Divide both sides by X(x)T(t): X''(x)/X(x) = (1/v^2)T''(t)/T(t) Since the left side depends only on x and the right side depends only on t, both sides must be equal to a constant. Let's call this constant -k^2: X''(x)/X(x) = -k^2 (1/v^2)T''(t)/T(t) = -k^2 Now we have two ordinary differential equations: X''(x) + k^2X(x) = 0 T''(t) + (v^2)k^2T(t) = 0 Solve the first equation for X(x): X(x) = A*sin(kx) + B*cos(kx) Apply the boundary conditions u(0, t) = 0 and u(L, t) = 0: X(0) = A*sin(0) + B*cos(0) = B = 0 X(L) = A*sin(kL) = 0 Since L = 1, we have: A*sin(k) = 0 Non-trivial solutions exist when k = n*pi, where n is an integer. So, X_n(x) = A_n*sin(n*pi*x). Now, solve the second equation for T(t): T(t) = C*cos(v*k*t) + D*sin(v*k*t) Apply the initial condition u_t(x, 0) = 0: T'(t) = -v*k*C*sin(v*k*t) + v*k*D*cos(v*k*t) T'(0) = D*v*k = 0 Since k ≠ 0, we have D = 0. So, T_n(t) = C_n*cos(n*pi*v*t). The general solution is: u(x, t) = sum(C_n*sin(n*pi*x)*cos(n*pi*v*t)) Apply the initial condition u(x, 0) = sin(pi*x): u(x, 0) = sum(C_n*sin(n*pi*x)) = sin(pi*x) Comparing the terms, we see that C_1 = 1 and C_n = 0 for n ≠ 1. Therefore, the solution to the given wave equation with the given initial and boundary conditions is: u(x, t) = sin(pi*x)*cos(pi*v*t)

answer:To solve this problem, we will use the method of separation of variables. We assume that the solution can be written in the form u(x,t) = X(x)T(t). Substituting this into the wave equation, we get: frac{partial^2 u}{partial t^2} = 4frac{partial^2 u}{partial x^2} Rightarrow X(x)T''(t) = 4X''(x)T(t) Now, we divide both sides by X(x)T(t): frac{T''(t)}{T(t)} = 4frac{X''(x)}{X(x)} Since the left side is a function of t only and the right side is a function of x only, both sides must be equal to a constant, which we will call -4lambda^2: frac{T''(t)}{T(t)} = -4lambda^2 = 4frac{X''(x)}{X(x)} Now we have two ordinary differential equations (ODEs) to solve: 1) T''(t) + 4lambda^2 T(t) = 0 2) X''(x) - lambda^2 X(x) = 0 The general solution for the second ODE is: X(x) = Acos(lambda x) + Bsin(lambda x) Since the string is fixed at both ends, we have the boundary conditions X(0) = X(1) = 0. Applying these conditions, we get: X(0) = Acos(0) + Bsin(0) = A = 0 X(1) = Bsin(lambda) = 0 Since B cannot be zero (otherwise the solution would be trivial), we must have sin(lambda) = 0. This implies that lambda = npi for n = 1, 2, 3, .... Thus, the solution for X(x) is: X_n(x) = B_nsin(npi x) Now, we solve the first ODE. The general solution is: T(t) = Ccos(2lambda t) + Dsin(2lambda t) Since the initial velocity is zero, we have frac{partial u}{partial t}(x,0) = X(x)T'(0) = 0. This implies that D = 0, and the solution for T(t) is: T_n(t) = C_ncos(2npi t) Now, the general solution for the wave equation is the sum of the product of these two solutions: u(x,t) = sum_{n=1}^{infty} C_ncos(2npi t)sin(npi x) To find the coefficients C_n, we use the initial displacement condition u(x,0) = frac{1}{2}x. This gives: frac{1}{2}x = sum_{n=1}^{infty} C_nsin(npi x) We can find the coefficients C_n by multiplying both sides by sin(mpi x) and integrating from 0 to 1: int_0^1 frac{1}{2}xsin(mpi x) dx = sum_{n=1}^{infty} C_nint_0^1 sin(npi x)sin(mpi x) dx Using the orthogonality property of sine functions, the right side simplifies to: C_mint_0^1 sin^2(mpi x) dx Now, we need to compute the left side integral: int_0^1 frac{1}{2}xsin(mpi x) dx We can solve this integral using integration by parts. Let u = x and dv = sin(mpi x) dx. Then, du = dx and v = -frac{1}{mpi}cos(mpi x). Applying integration by parts, we get: int_0^1 frac{1}{2}xsin(mpi x) dx = -frac{1}{2mpi}left[xcos(mpi x)right]_0^1 + frac{1}{2mpi}int_0^1 cos(mpi x) dx The second integral is zero, so we are left with: -frac{1}{2mpi}left[cos(mpi) - 0right] = frac{(-1)^{m+1}}{2mpi} Now, equating both sides, we get: C_mint_0^1 sin^2(mpi x) dx = frac{(-1)^{m+1}}{2mpi} The integral on the left side is frac{1}{2}, so we have: C_m = frac{(-1)^{m+1}}{mpi} Finally, the solution for the displacement of the string is: u(x,t) = sum_{n=1}^{infty} frac{(-1)^{n+1}}{npi}cos(2npi t)sin(npi x) For t = 2, the displacement is: u(x,2) = sum_{n=1}^{infty} frac{(-1)^{n+1}}{npi}cos(4npi)sin(npi x) = sum_{n=1}^{infty} frac{(-1)^{n+1}}{npi}sin(npi x) This is the displacement of the string for 0 leq x leq 1 and t = 2.