Physics - Standing Waves

Zsmart.id - In this topic, we will explore standing waves, including the definition of standing waves and the characteristics of standing waves on strings with free and fixed ends.

A. Definition of Standing Waves

A standing wave is a wave that forms due to the interference of two waves with the same frequency and amplitude, moving in opposite directions. This results in a stationary pattern with certain points remaining stationary (nodes) and other points oscillating with maximum amplitude (antinodes).

B. Equation of Standing Waves on a String with Free Ends

On a string with free ends, a standing wave is formed with antinodes (points with maximum amplitude) at the ends of the string. The equation for a standing wave on a string with free ends is:

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

Where:

• 2A is the amplitude of the standing wave at the free ends of the string (twice the amplitude of the traveling wave due to constructive interference).
• A is the amplitude of the traveling wave.
• k is the wave number, $k = \frac{2\pi}{\lambda}$, where $\lambda$ is the wavelength.
• x is the position along the string.
• ω is the angular frequency, $\omega = 2\pi f$, where $f$ is the frequency of the wave.
• t is time.

At the free ends (at position $x = L$, the length of the string), there are antinodes, where the maximum oscillation occurs.

C. Equation of Standing Waves on a String with Fixed Ends

For a string with both ends fixed, standing waves are formed with nodes (points that remain stationary) at both ends of the string. The equation for a standing wave on a string with fixed ends is:

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

Where:

• 2A is the amplitude of the standing wave on the string with fixed ends.
• k and ω have the same meanings as in the above equation.

At the fixed ends of the string, there are nodes, while the center of the string has antinodes.

D. Equations for Determining the Position of Antinodes (Perut) and Nodes (Simpul)

In standing waves, there are specific points called nodes and antinodes.

• Node: A point where the medium particles do not move at all (maximum displacement = 0). Destructive interference occurs between the two waves moving in opposite directions.
• Antinode: A point where the medium particles move with maximum amplitude. Constructive interference occurs between the two waves moving in opposite directions.

1. For a String with Fixed Ends

For a string with fixed ends, nodes are located at the ends of the string, and antinodes are located at specific points along the string.

• Nodes occur at:

  $$x_n=\frac{nL}{2}$$

  Where:

  • $n=0,1,2,3,\dots$ is the node number.
  • $L$ is the length of the string.

• Antinodes occur at:

  $$x_n=\frac{(2n+1)L}{4}$$

  Where:

  • $n=0,1,2,3,\dots$ is the antinode number.

2. For a String with Free Ends

For a string with free ends, antinodes are located at the ends of the string, and nodes are located at specific points along the string.

• Antinodes occur at:

  $$x_n=\frac{(2n+1)L}{2}$$

  Where:

  • $n=0,1,2,3,\dots$ is the antinode number.
  • $L$ is the length of the string.

• Nodes occur at:

  $$x_n=\frac{nL}{2}$$

  Where:

  • $n=1,2,3,\dots$ is the node number.

E. Frequency of Standing Waves on Strings with Free and Fixed Ends

The frequency of standing waves on a string depends on the length of the string $L$, the tension $T$, and the linear mass density $\mu$ of the string. The frequency equation for standing waves is:

$$f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$$

• n is the harmonic number (1 for the fundamental frequency, 2 for the second harmonic, and so on).
• L is the length of the string.
• T is the tension in the string.
• μ is the linear mass density of the string.

For free-end strings, the standing wave has one antinode in the center and nodes at the ends. For fixed-end strings, there are nodes at both ends and an antinode in the middle.

F. Example Problems

Problem 1: A string with a length of 1 meter and a linear mass density of 0.05 kg/m is stretched with a tension of 100 N. Calculate the fundamental frequency for the string with fixed ends.

To calculate the fundamental frequency of a string with fixed ends, we use the following formula:

$$f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$

Where:

• $f_1$ is the fundamental frequency,
• $L$ is the length of the string,
• $T$ is the tension in the string,
• $\mu$ is the linear mass density of the string.

Given values:

• $L = 1 \, \text{m}$
• $\mu = 0.05 \, \text{kg/m}$
• $T = 100 \, \text{N}$

Now, substitute the given values into the formula:

$$f_1 = \frac{1}{2 \times 1} \sqrt{\frac{100}{0.05}}$$

Simplify the expression inside the square root:

$$f_1 = \frac{1}{2} \sqrt{\frac{100}{0.05}} = \frac{1}{2} \sqrt{2000}$$

Now calculate $\sqrt{2000}$:

$$f_1=\frac{1}{2}\times 44.72=22.36\text{Hz}$$

The fundamental frequency $f_1$ for the string with fixed ends is 22.36 Hz.

Problem 2: A string with a free end, a length of 1.5 meters, and a linear mass density of 0.03 kg/m is under a tension of 80 N. Calculate the frequency for the third harmonic.

To calculate the frequency for the third harmonic of a string with a free end, we will use the following formula:

$$f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$$

Where:

• $f_n$ is the frequency for the $n$-th harmonic,
• $n$ is the harmonic number (3 in this case for the third harmonic),
• $L$ is the length of the string,
• $T$ is the tension in the string,
• $\mu$ is the linear mass density of the string.

Given values:

• $L = 1.5 \, \text{m}$
• $\mu = 0.03 \, \text{kg/m}$
• $T = 80 \, \text{N}$
• $n = 3$ (since we're looking for the third harmonic)

Now, substitute the given values into the formula:

$$f_3 = \frac{3}{2 \times 1.5} \sqrt{\frac{80}{0.03}}$$

Simplify the expression inside the square root:

$$f_3 = \frac{3}{3} \sqrt{\frac{80}{0.03}} = \sqrt{\frac{80}{0.03}} = \sqrt{2666.67}$$

Now calculate $\sqrt{2666.67}$:

$$f_3\approx 51.63\text{Hz}$$

The frequency for the third harmonic $f_3$ is approximately 51.63 Hz.

F.  Problems on Standing Waves

1. Problem 1: A string with a length of 3 meters and a linear mass density of 0.1 kg/m is stretched with a tension of 120 N. Determine the fundamental frequency for the string with fixed ends.
2. Problem 2: A string with a free end, a length of 1.5 meters and a linear mass density of 0.03 kg/m is under a tension of 80 N. Calculate the frequency for the third harmonic.
3. Problem 3: A string with a length of 4 meters has a linear mass density of 0.04 kg/m and is under a tension of 60 N. Determine the fundamental frequency and the frequency of the second harmonic for a string with a free end.
4. Problem 4: A string with a length of 2 meters is fixed at both ends, and it vibrates in the second harmonic mode. The tension in the string is 100 N, and the linear mass density of the string is 0.05 kg/m. Determine the positions of the 3rd nodes and 4th antinodes along the string. 

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