A string is stretched so that it is under tension and is tied at both ends so that the endpoints don't move. A mechanical oscillator then vibrates the string so that a standing wave is created. The dark line in each diagram represents a snapshot of a string at an instant in time when the amplitude of the standing wave is a maximum. The lighter lines represent the string at other times during a complete cycle. All of the strings have the same lengths but may not have the same mass. The number of nodes and antinodes in the standing wave is the same in Case A and Case D. The tensions in the strings(T) and the standing wave frequencies(f) are given in each figure. Rank the speed of the wave in the strings.Enter in the form A > B = C ... make sure you separate characters with a space. Equal values can be entered in either order. Hint: v=(T/μ)0.5. ### Comparative Analysis of Wave Properties Under Different Tensions and FrequenciesIn this section, we explore wave behavior under various tension (T) and frequency (f) conditions. Each of the four diagrams, labeled A, B, C, and D, illustrates a wave traveling under specific tension and frequency parameters. A table below illustrates their respective conditions:| Diagram | Tension (T) | Frequency (f) ||---------|--------------|---------------|| A | 2 N | 500 Hz || B | 5 N | 300 Hz || C | 6 N | 300 Hz || D | 2 N | 400 Hz |#### Diagram A: \( T = 2 \, \text{N} \), \( f = 500 \, \text{Hz} \)Here, the wave is plotted with a tension of 2 Newtons and a frequency of 500 Hertz. The wave shows several complete oscillations over the string, indicating a high frequency.#### Diagram B: \( T = 5 \, \text{N} \), \( f = 300 \, \text{Hz} \)This wave is under a tension of 5 Newtons with a frequency of 300 Hertz. There are fewer oscillations compared to Diagram A, suggesting a lower frequency.#### Diagram C: \( T = 6 \, \text{N} \), \( f = 300 \, \text{Hz} \)In this diagram, the wave’s tension is increased to 6 Newtons while maintaining the same frequency of 300 Hertz as Diagram B. The increase in tension results in a visibly different waveform.#### Diagram D: \( T = 2 \, \text{N} \), \( f = 400 \, \text{Hz} \)With a tension of 2 Newtons and a frequency of 400 Hertz, this diagram depicts more oscillations than those seen in Diagrams B and C but fewer than in Diagram A.### Analysis and ComparisonBy examining these diagrams, students can better understand the relationships between tension, frequency, and wave behavior. Higher frequencies result in more oscillations, while increased tension generally affects the wavelength and amplitude. Comparing Diagram A with Diagram D, both at the same tension (2 Newtons), the difference in frequency (500 Hz vs. 400 Hz) is evident

The frequency of the wave generated by the string fixed at both the ends is given by,

A string is stretched so that it is under tension and is tied at both ends so that the endpoints don't move. A mechanical oscillator then vibrates the string so that a standing wave is created. The dark line in each diagram represents a snapshot of a string at an instant in time when the amplitude of the standing wave is a maximum. The lighter lines represent the string at other times during a complete cycle. All of the strings have the same lengths but may not have the same mass. The number of nodes and antinodes in the standing wave is the same in Case A and Case D. The tensions in the strings(T) and the standing wave frequencies(f) are given in each figure. Rank the speed of the wave in the strings. Enter in the form A > B = C ... make sure you separate characters with a space. Equal values can be entered in either order. Hint: v=(T/μ)0.5.

A string is stretched so that it is under tension and is tied at both ends so that the endpoints don't move. A mechanical oscillator then vibrates the string so that a standing wave is created. The dark line in each diagram represents a snapshot of a string at an instant in time when the amplitude of the standing wave is a maximum. The lighter lines represent the string at other times during a complete cycle. All of the strings have the same lengths but may not have the same mass. The number of nodes and antinodes in the standing wave is the same in Case A and Case D. The tensions in the strings(T) and the standing wave frequencies(f) are given in each figure. Rank the speed of the wave in the strings. Enter in the form A > B = C ... make sure you separate characters with a space. Equal values can be entered in either order. Hint: v=(T/μ)0.5.