find the solution of the wave equation for a vibrating string of length (l)=10units with ends fixed .it is given that c^2=1,initial velocity is zero and initial deflection is u(x,o)=k(sinx-sin2x)
it i'll be 60m
as there is no external force acting except gravitational force ,do its path should be at the range of parabolic path
range(R)=u^2sin(theta)/g
here u=20,theta=45,g=10
thus R=20^2sin45/10=4
conservation of momentum m1d1+m2d2=md
m/2*20+m/2*d2=m40
d2=60
the other piece will fall 60m far from origin
To find the solution of the wave equation for a vibrating string with fixed ends, we'll use the method of separation of variables. The wave equation for a vibrating string can be written as:
∂²u/∂t² = c² ∂²u/∂x²
where u(x, t) represents the displacement of the string at position x and time t, and c represents the wave speed. Given that c² = 1 (which implies c = 1 since c cannot be negative), the wave equation simplifies to:
∂²u/∂t² = ∂²u/∂x²
Now, let's find the general solution by assuming that the displacement can be represented as a product of two functions:
u(x, t) = X(x)T(t)
Substituting this assumption into the wave equation, we get:
X''(x)T(t) = X(x)T''(t)
Dividing both sides by X(x)T(t), we have:
X''(x)/X(x) = T''(t)/T(t)
Since the left-hand side only depends on x and the right-hand side only depends on t, both sides must be equal to a constant value, which we'll call -λ². This gives us two separate ordinary differential equations to solve:
X''(x) = -λ²X(x)
T''(t) = -λ²T(t)
Solving the differential equation for X(x), we find that the general solution is:
X(x) = A sin(λx) + B cos(λx)
where A and B are constants to be determined. However, since the vibrating string has fixed ends, the displacement at the ends must be zero, which implies that X(0) = 0 and X(l) = 0. Using these boundary conditions, we can determine the values of A and B:
X(0) = 0: A sin(0) + B cos(0) = 0
B = 0
X(l) = 0: A sin(λl) = 0
sin(λl) = 0
For sin(λl) = 0 to hold, λl must be an integer multiple of π. Therefore, we have:
λl = nπ
where n is an integer. Solving for λ, we have:
λ = (nπ)/l
Substituting the value of λ back into the equation for X(x), we have:
X(x) = A sin((nπx)/l)
Now, let's solve the differential equation for T(t). We have:
T''(t) = -λ²T(t) = -((nπ)/l)²T(t)
T''(t) + ((nπ)/l)²T(t) = 0
The general solution to this differential equation is given by:
T(t) = C₁ cos((nπt)/l) + C₂ sin((nπt)/l)
Combining the solutions for X(x) and T(t), we get the general solution for u(x, t):
u(x, t) = Σ(Aₙ sin((nπx)/l) cos((nπt)/l) + Bₙ sin((nπx)/l) sin((nπt)/l))
where Σ represents the sum over all integer values of n. The constants Aₙ and Bₙ can be determined using the initial condition u(x, 0) = k(sinx - sin2x), where k is a constant representing the initial deflection.