Question

A string with linear mass density \(\mu=0.0250 \mathrm{~kg} / \mathrm{m}\) under a tension of \(F=250 . \mathrm{N}\) is oriented in the \(x\) -direction. Two transverse waves of equal amplitude and with a phase angle of zero (at \(t=0\) ) but with different angular frequencies \((\omega=3000 . \mathrm{rad} / \mathrm{s}\) and \(\omega / 3=1000 . \mathrm{rad} / \mathrm{s})\) are created in the string by an oscillator located at \(x=0 .\) The resulting waves, which travel in the positive \(x\) -direction, are reflected at a distant point where both waves have a node, so there is a similar pair of waves traveling in the negative \(x\) -direction. Find the values of \(x\) at which the first two nodes in the standing wave are produced by these four waves.

Short answer

Answer: The first two nodes in the standing wave are produced at \(x_1 = \frac{\pi}{15} \mathrm{m}\) and \(x_2 = \frac{\pi}{5} \mathrm{m}\).

Step-by-step solution

Write down the displacement equation of each wave in the positive x-direction

We will use the following wave equation for the displacement of a wave on the string: \(y(x, t) = A \sin(kx - \omega t + \phi)\) where A is the amplitude, k is the wave number, \(\omega\) is the angular frequency, x is the position on the string, t is the time, and \(\phi\) is the phase angle. For the first wave in the positive x-direction, we have: \(y_1(x, t) = A \sin(3000 t - k_1 x)\), as \(\phi = 0\) and \(\omega = 3000 \mathrm{rad} / \mathrm{s}\). For the second wave in the positive x-direction, we have: \(y_2(x, t) = A \sin(1000 t - k_2 x)\), as \(\phi = 0\) and \(\omega = 1000 \mathrm{rad} / \mathrm{s}\).

Calculate the wave number for each wave

We know that the wave speed for both waves is the same given by the formula: \(v = \sqrt{\frac{F}{\mu}}\) where F is the tension force and \(\mu\) is the linear mass density. \(k = \frac{\omega}{v}\) Calculating the wave speed, \(v = \sqrt{\frac{250 . \mathrm{N}}{0.0250 \mathrm{~kg} / \mathrm{m}}} = 100 \mathrm{m} / \mathrm{s}\) Now, let's find the wave number for each wave: \(k_1 = \frac{3000 \mathrm{rad} / \mathrm{s}}{100 \mathrm{m} / \mathrm{s}} = 30 \mathrm{m}^{-1}\) \(k_2 = \frac{1000 \mathrm{rad} / \mathrm{s}}{100 \mathrm{m} / \mathrm{s}} = 10 \mathrm{m}^{-1}\)

Write down the displacement equation of each wave in the negative x-direction

The reflected waves in the negative x-direction have the same amplitude and angular frequency as the original waves but will have a slightly different wave equation: For the first wave in the negative x-direction, we have: \(y_3(x, t) = A \sin(3000 t + k_1 x)\) For the second wave in the negative x-direction, we have: \(y_4(x, t) = A \sin(1000 t + k_2 x)\)

Write down the total displacement equation for the standing wave

Using the principle of superposition, the total displacement of the standing wave is: \(y(x, t) = y_1(x, t) + y_2(x, t) + y_3(x, t) + y_4(x, t)\) \(y(x, t) = A [\sin(3000 t - k_1 x) + \sin(3000 t + k_1 x) + \sin(1000 t - k_2 x) + \sin(1000 t + k_2 x)]\)

Find the nodes of the standing wave

At nodes, the total displacement of the standing wave must be zero: \(y(x, t) = 0\) Now we look for values of x that satisfy this equation for any time t. Notice that sine functions with different arguments, like \(3000t-k_1x\) and \(3000t+k_1x\), will cancel each other out since sin(x) = -sin(-x). Similarly for the other terms, they will also cancel each other. So, we only need to find the values of x at which the first two nodes occur. Since nodes happen when the sin functions become zero, we can write: For the first node: \(k_1x = n_1\pi: \frac{2 \pi}{k_1} = x_1\) \(x_1 = \frac{2 \pi}{30 \mathrm{m}^{-1}} = \frac{\pi}{15} \mathrm{m}\) For the second node: \(k_2x = n_2\pi: \frac{2 \pi}{k_2} = x_2\) \(x_2 = \frac{2 \pi}{10 \mathrm{m}^{-1}} = \frac{\pi}{5} \mathrm{m}\) So, the first two nodes in the standing wave are produced at \(x_1 = \frac{\pi}{15} \mathrm{m}\) and \(x_2 = \frac{\pi}{5} \mathrm{m}\).

Key concepts

Understanding Linear Mass Density

When studying waves, particularly those on strings, one fundamental property you'll encounter is linear mass density (\textmu). Think of linear mass density as how much mass exists within a given length of the string. It’s measured in kilograms per meter (kg/m) and has a direct impact on how waves behave on the string.

For instance, a string with a high linear mass density (heavier per unit length) requires more tension to achieve the same wave speed as a string with lower mass density. The equation tying tension (F), linear mass density (\textmu), and wave speed (v) together is given by: \[ v = \sqrt{\frac{F}{\textmu}} \]
This equation reveals that the wave speed is directly proportional to the square root of the tension and inversely proportional to the square root of the linear mass density. So, if we increase the mass density (making the string heavier), the wave will move slower, all else being equal. When it comes to standing waves, linear mass density contributes to determining the wave's velocity and ultimately where the nodes and antinodes (points of minimum and maximum wave displacement, respectively) will be located on a string.

Deciphering Angular Frequency

In the realm of wave motion, angular frequency (\textomega) offers insight into the 'speed' of the wave's oscillations—how quickly it cycles through its full range of motion. It’s expressed in radians per second (rad/s) and relates to the wave number (k) and the wave speed (v) through the relationship: \[ k = \frac{\textomega}{v} \]

Angular frequency is particularly important in discussions of wave interference and standing waves because it factors into the position and occurrence of nodes and antinodes. For example, a change in angular frequency while holding wave speed constant will result in a different wavelength and thus alter the pattern and position of nodes within standing waves. Let's not confuse angular frequency with the regular frequency, though—the latter simply tells us how many oscillations occur per second, whereas the former adds the 'angular' aspect, tying the concept to circular motion and waves.

Exploring Wave Interference

One of the most fascinating aspects of wave behavior is wave interference, which is when two or more waves pass through the same medium and overlap. This can occur with all types of waves, including sound, light, and, as in our exercise, waves on a string. The principle of superposition is a key rule for understanding wave interference, stating that the resulting wave displacement at any point is the sum of the displacements of individual waves at that point.

When waves interfere perfectly synchronously, they amplify each other, a phenomenon known as constructive interference. On the other hand, destructive interference occurs when waves are out of phase, canceling each other out and producing nodes—points of zero displacement. In our exercise, it's this destructive interference that leads us to determine the positions of the nodes in a standing wave. Standing waves arise when waves reflect and interfere with their own incoming waves. Think of it like an echo bouncing back and creating a pattern that seems to stand still. This is where we get to see those nodes (still points) and antinodes (points that oscillate with maximum amplitude) that define standing waves.