91Ó°ÊÓ

Standing waves are a fascinating phenomenon in physics that occur when waves are confined to a specific space and meet certain boundary conditions. On a string stretched between two supports, like in the example, the ends are fixed. These fixed ends act as nodes, or points where there is no movement, in the wave pattern. The nature of standing waves means that the wave's overall pattern remains stable over time, with certain points (antinodes) reaching maximum amplitude.

Each mode of vibration that a string can support is related to the concept of standing waves. For a string with fixed ends, an integral number of half-wavelengths can fit. This leads to discrete resonant modes which are critical for understanding vibrations in musical instruments and engineering structures.

• Nodes: Points along the string that remain stationary.
• Antinodes: Points where the amplitude of the standing wave reaches its peak.
• Boundary Conditions: For fixed ends, nodes must occur at the supports.

Calculating the wavelength in this context involves understanding how wavelengths fit into the string's length. Because standing waves form, we use the formula \(L = n(\lambda_{n} / 2)\). Here, \(L\) is the total length of the 2-meter string, \(n\) is the mode integer, and \(\lambda_{n}\) is the wavelength for that mode.

The relationship becomes apparent as we solve for \(\lambda_{n}\):\[ \lambda_{n} = \frac{2L}{n} \]
With \(n\) being the mode's number, starting from 1, 2, and upwards, the wavelength becomes smaller with each increasing mode. This is due to more half-wavelengths fitting into the same length. In this exercise:

• Mode 1: Wavelength is 4 m.
• Mode 2: Wavelength is 2 m.
• Mode 3: Wavelength is \(\frac{4}{3}\) m.

Understanding the reduction in wavelength as the mode number increases is essential in physics, as it affects how the wave interacts with its surroundings and the sound it produces.

Frequency tells us how many cycles of the wave pass a point per second. For a standing wave, you can find the frequency using the wave speed \(v_{w}\) and the wavelength \(\lambda_{n}\) through the formula:\[ v_{w} = f_{n} \lambda_{n} \]
To solve for frequency \(f_{n}\), rearrange to:\[ f_{n} = \frac{v_{w}}{\lambda_{n}} \]
Using the given wave speed of 50 m/s for our string, we calculated the frequencies for each mode:

• Mode 1: Frequency is 12.5 Hz, reflecting a larger wavelength, leading to a lower pitch.
• Mode 2: Frequency of 25 Hz, with everything aligning to half of Mode 1's wavelength.
• Mode 3: Frequency jumps to 37.5 Hz, as the shorter wavelength contributes to a higher pitch.

Both wavelength and frequency are fundamentally linked. As wavelength decreases, frequency proportionally increases, enabling an understanding of the dynamics involved in sound production and wave mechanics.