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.

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.

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: \(k_1 = \frac{\omega_1}{v} = \frac{3000}{100} = 30\, \text{rad/m}\) \(k_2 = \frac{\omega_2}{v} = \frac{1000}{100} = 10\, \text{rad/m}\)

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

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 \(x = 4.71\,\text{m}\).