Problem 72 - College Physics

Question

You are setting up two standing string waves. You have a length of uniform piano wire that is 3.0 m long and has a mass of 0.150 kg. You cut this into two lengths, one of 1.0 m and the other of 2.0 m, and place each length under tension. What should be the ratio of tensions (expressed as short to long) so that their fundamental frequencies are the same?

Step-by-Step Solution

Answer

The ratio of tensions should be 1:4.

Step 1: Understand the Problem

We have two pieces of piano wire, one that is 1.0 m long and the other that is 2.0 m long. Each piece is set up as a string with a standing wave, and we need to find the ratio of tensions in the two wires that will ensure both pieces have the same fundamental frequency.

Step 2: Formula for Fundamental Frequency

The fundamental frequency for a string is given by \( f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} \), where \( f \) is the frequency, \( L \) is the length of the string, \( T \) is the tension, and \( \mu \) is the linear mass density. For the frequencies to be equal, the equations for each string must equate.

Step 3: Determine Linear Mass Density

Linear mass density, \( \mu \), is defined as mass per unit length. For the 1.0 m string, \( \mu_1 = \frac{0.150 \, \text{kg}}{3.0 \, \text{m}} = 0.050 \, \text{kg/m} \). Since the mass is proportional to the length (and the distribution is uniform), \( \mu_2 = \mu_1 = 0.050 \, \text{kg/m} \).

Step 4: Equalize the Frequencies

Set the fundamental frequency formulas equal: \[ \frac{1}{2L_1}\sqrt{\frac{T_1}{\mu_1}} = \frac{1}{2L_2}\sqrt{\frac{T_2}{\mu_2}} \]. This simplifies to \[ L_1^2 \cdot T_2 = L_2^2 \cdot T_1 \] when considering equal \( \mu \).

Step 5: Solve for Tension Ratio

Substitute the lengths: \( L_1 = 1.0 \) m and \( L_2 = 2.0 \) m into \( L_1^2 \cdot T_2 = L_2^2 \cdot T_1 \), giving us \( 1.0^2 \cdot T_2 = 2.0^2 \cdot T_1 \), leading to \( T_2 = 4 \cdot T_1 \). Thus, \( \frac{T_1}{T_2} = \frac{1}{4} \).

Step 6: Express the Ratio

The ratio of the tensions (short string to long string) should be \( 1:4 \).

Key Concepts

Standing Waves

Standing waves are a fascinating phenomenon in wave physics, often observed on strings and other mediums. A standing wave occurs when two waves of the same frequency and amplitude, traveling in opposite directions, interfere with each other. This interference creates nodes and antinodes.
- **Nodes**: Points where the displacement is always zero. These are the quiet spots. - **Antinodes**: Points of maximum displacement. These spots "move" the most.
A characteristic feature of standing waves on strings is they can only form at certain frequencies. These are called the fundamental frequency and its harmonics. The fundamental frequency is the lowest frequency at which a standing wave can form. This is important because this frequency depends on the properties of the string, like length and tension.

Frequency Equation

The frequency of a standing wave on a string is determined by several factors and can be calculated using the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Where:

• \( f \) is the frequency.
• \( L \) is the length of the string.
• \( T \) is the tension in the string.
• \( \mu \) is the linear mass density.

This formula tells us that the frequency is inversely proportional to the length of the string. This means as the string's length increases, the frequency decreases. Conversely, frequency increases with the square root of the tension, highlighting the dependency of frequency on the tension applied to the string. This relationship is crucial for solutions involving setting up experiments with strings and waves, ensuring precise control over the resulting sound or wave pattern.

Linear Mass Density

Linear mass density, symbolized as \( \mu \), describes how mass is distributed along a string. It is calculated as mass per unit length:
\[ \mu = \frac{m}{L} \]
Where:

• \( m \) is the mass of the string.
• \( L \) is the total length of the string.

In the example, a 3.0 m long piano wire weighs 0.150 kg, giving a uniform linear mass density since the wire is divided equally. This distribution remains consistent across different string lengths. Knowing \( \mu \) is vital as it directly impacts the wave properties, particularly when calculating the frequency as seen in the frequency equation. It essentially sets the foundation for how waves travel along a medium.

Tension in Strings

Tension is the force applied along the length of a string. It plays a critical role in wave physics, particularly for standing waves. When a string is under tension, it becomes capable of supporting wave movements, and its tension directly influences the wave speed, and therefore the frequency.
In the exercise, students explored the concept of changing tension to achieve equal frequencies on strings of different lengths. To ensure that waves on both 1.0 m and 2.0 m strings have the same fundamental frequency, the tension ratio needed to be found. This comes from balancing the derived equation:
\[ L_1^2 \cdot T_2 = L_2^2 \cdot T_1 \]
Through this, it is understood that the tension in the shorter string must be less. The solution derived that for the strings to have the same frequency, the ratio of tension \( T_1:T_2 \) needs to be \( 1:4 \). This highlights how adjusting tension is a method for controlling wave behavior in practical applications such as musical instruments or physics experiments.