Problem 13 - Physics

Question

The middle \(C\) string on a piano is under a tension of \(944 \mathrm{~N}\). The period and wavelength of a wave on this string are \(3.82 \mathrm{~ms}\) and \(1.26 \mathrm{~m}\), respectively. Find the linear density of the string.

Step-by-Step Solution

Answer

The linear density of the string is approximately \(0.00153 \, \text{kg/m}\).

Step 1: Identify the Given Values

We are given a tension \(T = 944 \, \text{N}\), period \(T = 3.82 \, \text{ms} = 3.82 \times 10^{-3} \, \text{s}\), and wavelength \(\lambda = 1.26 \, \text{m}\).

Step 2: Calculate the Frequency

The frequency \(f\) is the reciprocal of the period. Thus, \(f = \frac{1}{T} = \frac{1}{3.82 \times 10^{-3}} \, \text{Hz}\). Calculate this to find \(f\).

Step 3: Calculate the Wave Speed

The wave speed \(v\) on the string is given by \(v = f \times \lambda\). Use the frequency from Step 2 and the wavelength to find \(v\).

Step 4: Use the Wave Equation for Strings

For a string, the wave speed \(v\) is also given by \(v = \sqrt{\frac{T}{\mu}}\), where \(\mu\) is the linear density. Use the known tension and the wave speed from Step 3 to solve for \(\mu\): \(\mu = \frac{T}{v^2}\).

Step 5: Final Calculation

Perform the calculation: Evaluate \(\mu = \frac{T}{v^2}\) to find the linear density of the string.

Key Concepts

Wave Speed Calculation

Understanding wave speed on a string is crucial to solving many physics problems. Wave speed (\(v\)) is an indicator of how fast a wave travels down a string or any medium. In this context, to determine the wave speed on a string, we use the formula:

• \( v = f \times \lambda \)

where \(f\) is frequency and \(\lambda\) is wavelength. Frequency is how many cycles occur in one second, and wavelength is the distance between two consecutive points in a wave.

Once you know the frequency and wavelength, calculating wave speed becomes simple. After determining frequency, multiply it by the wavelength. Students frequently encounter this relationship when dealing with waves in physics. Hence, always having these concepts ready can make tackling wave problems straightforward.

Wave Equation for Strings

The physics of waves on a string can be beautifully encapsulated by the wave equation specific to strings, which is:

• \(v = \sqrt{\frac{T}{\mu}}\)

This equation shows wave speed \(v\) as a function of tension \(T\) in the string and the string's linear density \(\mu\). Linear density is the mass per unit length of the string.

Understanding this equation helps us see how

• Increasing tension (\(T\)) results in higher wave speeds
• Higher density (\(\mu\)) leads to slower wave speeds

If you have calculated the wave speed using the frequency and wavelength, you can use this equation to find the linear density of the string. Rearranging the equation yields:

• \(\mu = \frac{T}{v^2}\)

This formula provides a direct link between the tension in a string and its properties, helping solve for unknown variables when others are given.

Frequency and Period Relationship

The relationship between frequency and period is a fundamental concept in physics. Simply put, the frequency (\(f\)) is the reciprocal of the period (\(T\)). The period is the time taken for one complete cycle of a wave. Mathematically described as:

• \(f = \frac{1}{T}\)

This means that if you know the period, you can easily determine the frequency, and vice versa. If a wave has a shorter period, it means more cycles occur in a second, and hence, the frequency is higher.

Similarly, if the period is longer, fewer cycles occur, resulting in a lower frequency. This relationship is straightforward yet crucial for solving physics problems involving waves, like determining the wave speed or any calculations related to wave motions in strings. Grasping this concept simplifies understanding how waves behave and interact in different environments.