Question

Find the three lowest frequencies of oscillation of a string of length \(L\), tension \(T\), and mass per unit length \(\rho\), if a mass \(m\) is fastened to the string a distance \(L / 4\) from one end. $$ \frac{\partial^{2} y}{\partial x^{2}}-\frac{\rho}{T} \frac{\partial^{2} y}{\partial t^{2}}=0 \quad(m \ll \rho L) $$

Short answer

Solve the transcendental equation \( \tan\left(\frac{ \omega L }{ 4 v } \right) > - \omega L m / 4T \) to find the the three lowest frequencies formally.

Step-by-step solution

Understand the wave equation for a string

The given partial differential equation \[ \frac{\frac{\rho}{T} \frac{\frac{y}{\frac{\frac{\frac{x}{4}}}}}}\] describes the small oscillations of a string under tension.

Set the boundary conditions

The string is fixed at both ends, so the boundary conditions are:y(0, t) = y(L, t) = 0. Additionally, the mass m at a distance \(L/4\) introduces a discontinuity in the slope of the string.

General Solution

The general solution to the wave equation for a string with fixed ends is:\[ y(x,t) = \sum_{n=1}^{\infty} \left( A_n \cos(\omega_n t) + B_n \sin(\omega_n t) \right) \sin\left(\frac{n\pi x}{L}\right) \]where \( \omega_n = \frac{n\pi}{L} \sqrt{\frac{T}{\rho}} \) are the natural frequencies.

Apply Discontinuity Condition

Apply the condition at \( x = L/4 \) due to the mass m attached. This gives: T\( \left. \frac{ \partial y }{ \partial x } \right\vert_{x=L/4^+} - \left. \frac{ \partial y }{ \partial x } \right\vert_{x=L/4^-} \) = -m \frac{ \partial^2 y }{ \partial t^2 } |_{x=L/4}

Solve for the Frequencies

The equation from the previous step results in additional conditions that lead to the transcendental equation governing the allowed frequencies:\( \tan\left(\frac{ \omega L }{ 4 v } \right) > - \omega L m / 4T \).Solve this transcendental equation numerically to find the lowest frequencies \(\omega_1, \omega_2, \omega_3\).

Key concepts

Wave Equation

The wave equation describes how waves, such as oscillations in a string, propagate. For a string of length \(L\), tension \(T\), and mass per unit length \(\rho\), the wave equation is given by:
\[ \frac{\frac{\rho}{T} \frac{\frac{\partial^2 y}{\frac{\frac{\frac{\partial^2 y}{\frac{\frac{x^2}{4}}}}}}}} \]
This expresses the relationship between the second spatial derivative of the string's displacement \(y\) with respect to position \(x\) and the second time derivative of \(y\). Essentially, it tells us that the acceleration of a point on the string is proportional to the curvature of the string at that point.

Boundary Conditions

Boundary conditions describe the constraints placed on the string's motion. For a string fixed at both ends, these conditions are: y(0, t) = 0 and y(L, t) = 0.
This means that the displacement \(y\) of the string is zero at both ends (\(x = 0\) and \(x=L\)).
In our problem, there is also an additional condition due to the mass \(m\) attached at \(x = L/4\). This mass introduces a discontinuity in the slope of the string, which adds complexity to the solution.

Natural Frequencies

Natural frequencies are the frequencies at which the string naturally oscillates. For a string with fixed ends, the general solution to the wave equation can be written as:
\[ y(x,t) = \sum_{n=1}^{\infty} \( A_n \cos(\omega_n t) + B_n \sin(\omega_n t) \) \sin\left(\frac{n\pi x}{L}\right) \]
Here, \( \omega_n = \frac{n\pi}{L} \sqrt{\frac{T}{\rho}} \) are the natural frequencies.
These frequencies depend on the tension \(T\), mass per unit length \(\rho\), and the length \(L\) of the string, and correspond to the normal modes of vibration of the string.

Oscillations

Oscillations refer to the repetitive variations in displacement about an equilibrium position. For a string, the oscillations are determined by the initial disturbance and boundary conditions.
The general solution involving sine and cosine terms indicates that the string's displacement \(y\) oscillates harmonically with time and spatially along the length of the string.
The presence of the mass \(m\) at \(x = L/4\) modifies these oscillations by introducing an additional constraint through the discontinuity in the string's slope at that position.

Transcendental Equation

A transcendental equation is one that cannot be solved algebraically for the frequencies. In this problem, the mass \(m\) at \(x = L/4\) leads to additional conditions expressed through the slope discontinuity:
\[ T\left. \frac{ \partial y }{ \partial x } \right\vert_{x=L/4^+} - \left. \frac{ \partial y }{ \partial x } \right\vert_{x=L/4^-} = -m \frac{ \partial^2 y }{ \partial t^2 } |_{x=L/4} \]
This condition results in the transcendental equation:
\[ \tan\left(\frac{ \omega L }{ 4 v } \right) = - \frac{ \omega L m }{ 4T } \]
Solving this equation numerically gives us the lowest frequencies \( \omega_1, \omega_2, \omega_3\). This is necessary because an explicit solution for \(\omega\) cannot be derived algebraically.