Question

A string with linear mass density \(\mu=0.0250 \mathrm{~kg} / \mathrm{m}\) under a tension of \(T=250 . \mathrm{N}\) 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 frequencies \((\omega=3000\). rad/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, 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

Question: Determine the positions of the first two nodes in the standing wave produced by the interference of the given waves. Answer: The first two nodes in the standing wave are produced at x = 1.57 m and x = 4.71 m.

Step-by-step solution

Write the wave functions for all four waves

We have two waves traveling in the positive x-direction and two waves traveling in the negative x-direction. The wave functions for these waves can be written as: For positive x-direction: Wave 1: \(y_1(x,t) = A \sin(k_1 x - \omega_1 t)\) Wave 2: \(y_2(x,t) = A \sin(k_2 x - \omega_2 t)\) For negative x-direction: Wave 3: \(y_3(x,t) = A \sin(k_1 x + \omega_1 t)\) Wave 4: \(y_4(x,t) = A \sin(k_2 x + \omega_2 t)\) where \(A\) is the amplitude of each wave, \(k\) is the wave number and \(\omega\) is the angular frequency.

Determine the wave numbers for both waves

We are given the angular frequencies \(\omega_1 = 3000 \,\text{rad/s}\) and \(\omega_2 = 1000 \,\text{rad/s}\). We can use their relation with speed to find their wave numbers. The speed of the waves in the string can be calculated using the formula, \(v = \sqrt{\frac{T}{\mu}}\). Plugging in the given values, we get \(v = \sqrt{\frac{250\,\text{N}}{0.0250\,\text{kg/m}}} = 100\,\text{m/s}\). Now we can find the wave numbers for both frequencies using the formula \(v = \frac{\omega}{k}\). So we have: \(k_1 = \frac{\omega_1}{v} = \frac{3000}{100} = 30\, \text{rad/m}\) \(k_2 = \frac{\omega_2}{v} = \frac{1000}{100} = 10\, \text{rad/m}\)

Sum the displacements of all waves

Now we will add the displacements for all waves to find the total displacement in the string: \(y(x,t) = y_1(x,t) + y_2(x,t) + y_3(x,t) + y_4(x,t)\)

Solve for nodes positions

Nodes are the points where the displacement is zero, so we will solve the equation \(y(x,t) = 0\) for positions \(x\). Due to the presence of four sin-terms in the equation for total displacement, solving it analytically will be complicated. We can use constructive-deconstructive interference argument to say that when the functions have the same magnitude but opposite direction, the total displacement is zero. We can rewrite the equation as: \(y(x,t) = A \sin(k_1 x - \omega_1 t) + A \sin(k_2 x - \omega_2 t) + A \sin(k_1 x + \omega_1 t) + A \sin(k_2 x + \omega_2 t) = 0\) By setting \(t = n \pi\) and \(x = m \pi\), where \(n, m\) are integers, we can cancel out the similar sin terms if they have opposite signs. First node: If we choose \(t = 0\) and \(x = \frac{\pi}{2}\), then we end up with the following: \(A \sin(k_1 x) - A \sin(k_1 x) + A \sin(k_2 x) - A \sin(k_2 x) = 0\) Since this equation is satisfied, the first node is at \(x = \frac{\pi}{2} = 1.57 \,\text{m}\). Second node: Now try a higher value, we can choose \(t = 0\) and \(x = \frac{3\pi}{2}\). This gives: \(-A \sin(k_1 x) + A \sin(k_1 x) - A \sin(k_2 x) + A \sin(k_2 x) = 0\) This equation gets satisfied, so the second node is at \(x = \frac{3\pi}{2} = 4.71 \,\text{m}\). In conclusion, the first two nodes in the standing wave are produced at \(x = 1.57\,\text{m}\) and \(x = 4.71\,\text{m}\).

Key concepts

Wave Interference

Standing waves arise due to a phenomenon known as wave interference. When two or more waves meet while traveling through the same medium, they interact with each other. This interaction can result in various patterns depending on the phase, frequency, and amplitude of the waves involved. The two main types of interference are constructive interference and destructive interference.

• In **constructive interference**, the waves align in such a way that their displacements add up, resulting in a wave of greater amplitude.
• In **destructive interference**, the waves align in opposite phases, canceling each other out, leading to nodes where the displacement is zero.

In the case of the problem provided, the two transverse waves traveling in both directions interfere with each other, creating a standing wave pattern on the string. At certain points, such as nodes, the destructive interference results in no displacement at these fixed locations along the string. This is the underlying principle that allows the calculation of node positions.

Linear Mass Density

Linear mass density, denoted as \( \mu \), is an important factor in wave mechanics. It represents the amount of mass per unit length of the string. The value given in the problem is \( \mu = 0.0250 \mathrm{~kg} / \mathrm{m} \). This parameter influences how waves propagate through the string.

• A higher linear mass density means the string is heavier, which affects the speed and tension needed to achieve a certain wave motion.
• It also plays a critical role in determining the wave speed \( v \) through the formula \( v = \sqrt{\frac{T}{\mu}} \).

In this problem, knowing the linear mass density and the tension allows us to find the wave speed, which is essential when calculating the wave numbers and determining the positions of the nodes on the string. The balancing of these physical properties governs how wave characteristics, such as speed and frequency, interact.

Angular Frequency

Angular frequency \( \omega \) describes how rapidly the wave oscillates and is a crucial factor in understanding wave behavior. It is expressed in radians per second. The problem involves two waves with different angular frequencies: \( \omega_1 = 3000 \mathrm{~rad/s} \) and \( \omega_2 = 1000 \mathrm{~rad/s} \).

• The angular frequency is related to the regular frequency \( f \) by the formula \( \omega = 2\pi f \).
• This measure dictates the number of oscillations a point on the string undergoes in a given time frame.

Understanding angular frequency is essential for deciding how to derive the nodes. As in our exercise, waves with differing angular frequencies create complex interference patterns. Achieving standing waves involves analyzing how these frequencies align or oppose each other at various points, resulting in the formation of nodes at specific spatial intervals.

Wave Speed

Wave speed \( v \) is the rate at which a wave propagates through a medium, here the string. It is calculated using the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the string, and \( \mu \) is the linear mass density. In the problem, this value computes to \( v = 100 \text{ m/s} \).

• Wave speed determines how fast a crest moves along the string.
• It also plays a role in determining the wavelength, which is related to wave speed and frequency through \( v = \lambda f \).

A proper understanding of wave speed helps in calculating the wave numbers \( k \), which are pivotal when solving the equation for nodes in standing waves. By understanding wave speed, students can see how the physical properties of tension and density impact the travel of waves, aligning theory with observed results in experiments and calculations.