The boundary value problem $T \frac { d ^ { 2 } y } { d x ^ { 2 } } + \rho \omega ^ { 2 } = 0 , y ( 0 ) = 0 , y ( L ) = 0$ is a model of the shape of a rotating string. Suppose $T$ and $\rho$ are constants. The critical angular rotation speed $\omega = \omega _ { n }$ , for which there exist non-trivial solutions are
A) $\omega _ { n } = ( T / \rho ) \left( \frac { n \pi } { L } \right) ^ { 2 }$
B) $\omega _ { n } = \sqrt { T / \rho } \frac { n \pi } { L }$
C) $\omega _ { n } = \sqrt { T / \rho } \frac { n \pi } { 2 L }$
D) $\omega _ { n } = ( T / \rho ) \left( \frac { n \pi } { 2 L } \right) ^ { 2 }$
E) $\omega _ { n } = \sqrt { \rho / T } \frac { n \pi } { L }$

Question

Quizplus · Accepted Answer

The differential equation given is a standard form for the vibration of strings, and the solution involves finding the eigenvalues $\omega_n$ that satisfy the boundary conditions. The correct formula for the critical angular rotation speed $\omega_n$ that results in non-trivial solutions (standing wave patterns) is given by $\omega_n = \sqrt{T/\rho} \frac{n\pi}{L}$, where $n$ is a positive integer representing the mode number, $T$ is the tension in the string, $\rho$ is the linear mass density of the string, and $L$ is the length of the string. This formula is derived by solving the differential equation with the given boundary conditions and represents the natural frequencies of the string's vibration.

The Boundary Value Problem Td2ydx2+ρω2=0,y(0)=0,y(L)=0T \frac { d ^ { 2 } y } { d x ^ { 2 } } + \rho \omega ^ { 2 } = 0 , y ( 0 ) = 0 , y ( L ) = 0Tdx2d2y​+ρω2=0,y(0)=0,y(L)=0

The Boundary Value Problem $T \frac { d ^ { 2 } y } { d x ^ { 2 } } + \rho \omega ^ { 2 } = 0 , y ( 0 ) = 0 , y ( L ) = 0$