Question:

A string of length 1.5 m and linear density 0.02 kg/m is fixed at both ends. Two waves are traveling on this string in opposite directions with frequencies of 100 Hz and 150 Hz respectively. The equation for the waves are given by:

Wave 1:

$$y_1 = 0.04 \sin(2\pi(100)t - 2\pi\frac{x}{1.5})$$

Wave 2:

$$y_2 = 0.06 \sin(2\pi(150)t - 2\pi\frac{x}{1.5})$$

where $y$ is the displacement of the string in meters, $t$ is time in seconds, and $x$ is position along the string in meters.

(a) Determine the interference pattern created by the two waves. Identify any nodes and antinodes and calculate their positions along the string.

(b) Determine the wavelength of the resulting standing wave and the distance between adjacent nodes.

(c) Calculate the speed at which the standing wave pattern appears to move along the string.

(d) Determine the equation for the resulting standing wave pattern.

(e) Calculate the amplitude of the resulting standing wave.

(f) Determine the maximum speed of a point on the string.

Answer:

(a) To determine the interference pattern created by the two waves, we need to superpose the waves and find the resultant amplitude at each point. Mathematically, this can be done by adding the two wave equations together:

$$y = y_1 + y_2$$

Substituting the given wave equations:

$$y = 0.04 \sin(2\pi(100)t - 2\pi\frac{x}{1.5}) + 0.06 \sin(2\pi(150)t - 2\pi\frac{x}{1.5})$$

Simplifying the equation:

$$y = 0.04 \sin(2\pi(100)t - 2\pi\frac{x}{1.5}) + 0.06 \sin(2\pi(150)t - 2\pi\frac{x}{1.5})$$

$$y = 0.04 \sin(200\pi t - \frac{400\pi x}{1.5}) + 0.06 \sin(300\pi t - \frac{600\pi x}{1.5})$$

To find the nodes and antinodes, we can observe that nodes occur when the resultant amplitude is zero, and antinodes occur when the resultant amplitude is maximum.

When the two waves have equal amplitudes and are in phase, the resultant amplitude is zero (node) at $t = 0$. Therefore, we have:

$$0.04 \sin(200\pi(0) - \frac{400\pi x}{1.5}) + 0.06 \sin(300\pi(0) - \frac{600\pi x}{1.5}) = 0$$

Simplifying the equation:

$$0.04 \sin( - \frac{400\pi x}{1.5}) + 0.06 \sin( - \frac{600\pi x}{1.5}) = 0$$

Using the trigonometric identity $\sin(-\theta) = -\sin(\theta)$, we can rewrite the equation as:

$$-0.04 \sin(\frac{400\pi x}{1.5}) - 0.06 \sin(\frac{600\pi x}{1.5}) = 0$$

$$-0.04 \sin(\frac{400\pi x}{1.5}) = 0.06 \sin(\frac{600\pi x}{1.5})$$

$$\sin(\frac{400\pi x}{1.5}) = -\frac{0.06}{0.04} \sin(\frac{600\pi x}{1.5})$$

$$\sin(\frac{400\pi x}{1.5}) = -1.5 \sin(\frac{600\pi x}{1.5})$$

Using the trigonometric identity $\sin(\theta) = \sin(\pi - \theta)$, we can rewrite the equation as:

$$\sin(\frac{400\pi x}{1.5}) = 1.5 \sin(\frac{\pi}{2} - \frac{600\pi x}{1.5})$$

Setting the arguments of sin equal to each other:

$$\frac{400\pi x}{1.5} = \frac{\pi}{2} - \frac{600\pi x}{1.5}$$

Simplifying the equation:

$$400\pi x = \frac{\pi}{2} - 600\pi x$$

$$1000\pi x = \frac{\pi}{2}$$

$$x = \frac{1}{2000} \approx 0.0005 \, \text{m}$$

Therefore, the position at which a node occurs is approximately 0.0005 m.

Similarly, we can find the position at which an antinode occurs by noticing that an antinode is formed when the two waves are completely out of phase. This occurs at $t = 0.0025$ s. We can calculate the position as follows:

$$0.04 \sin(200\pi(0.0025) - \frac{400\pi x}{1.5}) + 0.06 \sin(300\pi(0.0025) - \frac{600\pi x}{1.5}) = 0.1$$

Simplifying the equation, we find:

$$\frac{400\pi x}{1.5} = \frac{\pi}{2} - \frac{600\pi x}{1.5}$$

$$800\pi x = \frac{\pi}{2} - 900\pi x$$

$$1700\pi x = \frac{\pi}{2}$$

$$x \approx 0.0009 \, \text{m}$$

Therefore, the position at which an antinode occurs is approximately 0.0009 m.

(b) The wavelength of the resulting standing wave can be determined from the distance between two adjacent nodes. We found that the distance between two nodes is approximately 0.0005 m. Since there is half a wavelength between adjacent nodes, the wavelength can be calculated as:

$$\lambda = 2 \times 0.0005 = 0.001 \, \text{m}$$

Therefore, the wavelength of the resulting standing wave is 0.001 m.

(c) The speed at which the standing wave pattern appears to move along the string can be calculated using the formula:

$$v = f \lambda$$

where $v$ is the speed, $f$ is the frequency, and $\lambda$ is the wavelength.

The frequency of the waves are given as 100 Hz and 150 Hz respectively. Substituting the values into the formula:

$$v = (100 + 150) \times 0.001 = 0.25 \, \text{m/s}$$

Therefore, the speed at which the standing wave pattern appears to move along the string is 0.25 m/s.

(d) The equation for the resulting standing wave pattern can be determined by observing that the waves have equal amplitudes and are 180 degrees out of phase. The equation for a standing wave with these conditions is given by:

$$y = 2A\sin(kx)\cos(\omega t)$$

where $A$ is the amplitude, $k$ is the wave number, $x$ is the position, and $\omega$ is the angular frequency.

Comparing this equation with the given wave equations, we can identify that the amplitude $A$ is equal to 0.04 m. Therefore, the equation for the resulting standing wave pattern is:

$$y = 2 \times 0.04 \sin\left(\frac{2\pi}{0.001} x\right)\cos(2\pi(100)t)$$

$$y \approx 0.08 \sin(2000\pi x)\cos(200\pi t)$$

Therefore, the equation for the resulting standing wave pattern is approximately $y \approx 0.08 \sin(2000\pi x)\cos(200\pi t)$.

(e) The amplitude of the resulting standing wave can be determined by looking at the given wave equations. The maximum amplitude between the two waves is 0.06 m. Therefore, the amplitude of the resulting standing wave is 0.06 m.

(f) The maximum speed of a point on the string can be determined by finding the maximum value of the velocity function. The velocity function for a standing wave is given by:

$$v = \omega A\cos(kx)\sin(\omega t)$$

where $\omega$ is the angular frequency, $A$ is the amplitude, $k$ is the wave number, $x$ is the position, and $t$ is the time.

Substituting the given values into the formula:

$$v = 2\pi(100) \times 0.06 \cos\left(\frac{2\pi}{0.001} x\right)\sin(2\pi(100)t)$$

Simplifying the equation:

$$v = 0.12\pi \cos(2000\pi x)\sin(200\pi t)$$

Therefore, the maximum speed of a point on the string is 0.12π m/s.

This concludes the solution for the given AP Physics 2 exam question on interference and standing waves.