Problem 39 - University Physics with Modern Physics

Question

A 2.00 -m-long string of mass \(10.0 \mathrm{~g}\) is clamped at both ends. The tension in the string is \(150 \mathrm{~N}\). a) What is the speed of a wave on this string? b) The string is plucked so that it oscillates. What is the wavelength and frequency of the resulting wave if it produces a standing wave with two antinodes?

Step-by-Step Solution

Answer

Based on the given information, the speed of the wave on the string is 173 m/s, and the wavelength and frequency of the standing wave with two antinodes are 2 meters and 86.5 Hz, respectively.

Step 1: Calculate the wave speed

To find the speed of a wave on the string, we can use the formula: Wave speed (v) = \(\sqrt{\frac{T}{\mu}}\) where T is the tension in the string and \(\mu\) is the linear mass density of the string, defined as the mass per unit length. First, we need to find the linear mass density. This can be calculated using: \(\mu\) = \(\frac{m}{L}\) where m is the mass of the string, and L is the length of the string.

Step 2: Calculate the linear mass density

We are given the mass of the string is 10g, and it is 2 meters long. So the linear mass density can be calculated as: \(\mu\) = \(\frac{10 \text{g}}{2 \text{m}}\) = \(\frac{10 \times 10^{-3} \text{kg}}{2 \text{m}}\) = \(5\times10^{-3} \frac{\text{kg}}{\text{m}}\)

Step 3: Calculate the wave speed

Now that we have the linear mass density, we can plug the values of tension and \(\mu\) into the wave speed formula: v = \(\sqrt{\frac{150 \, \text{N}}{5\times10^{-3} \, \frac{\text{kg}}{\text{m}}}}\) v = \(\sqrt{30,000 \, \frac{\text{m}^2}{\text{s}^2}} = 173 \, \frac{\text{m}}{\text{s}}\) The wave speed on this string is 173 m/s.

Step 4: Determine the wavelength of the standing wave

Now that we have the wave speed, we can move on to finding the wavelength of the standing wave with two antinodes. For a standing wave with two antinodes, the string length is equal to one full wavelength. So, the wavelength (\(\lambda\)) is: \(\lambda = 2 \text{m}\)

Step 5: Determine the frequency of the standing wave

Finally, we can calculate the frequency of the standing wave using the wave speed, v, and the wavelength, \(\lambda\): Frequency (f) = \(\frac{v}{\lambda}\) f = \(\frac{173 \, \frac{\text{m}}{\text{s}}}{2 \, \text{m}}\) = \(86.5 \text{ Hz}\) The frequency of the standing wave with two antinodes is 86.5 Hz. In summary, the speed of a wave on the string is 173 m/s, and the wavelength and frequency of the resulting standing wave with two antinodes are 2 meters and 86.5 Hz, respectively.

Key Concepts

Wave Speed

The speed of a wave on a string is a fundamental concept in wave mechanics.
Wave speed determines how quickly the wave travels along the string.
It is crucial in understanding wave behavior in different mediums. To find the wave speed, you can use the formula:

• Wave speed (v) = \( \sqrt{\frac{T}{\mu}} \)

Here:

• \( T \) is the tension applied to the string.
• \( \mu \) is the linear mass density of the string.

These two factors influence the speed at which waves propagate.
Essentially, greater tension and a lower linear mass density lead to faster wave speeds. In the given problem, the calculation showed a wave speed of 173 m/s.

Standing Wave

Standing waves are a fascinating phenomenon in wave mechanics, occurring when two waves of the same frequency and amplitude travel in opposite directions and interfere.
They give the appearance of a wave that is stationary, with nodes (points of no motion) and antinodes (points of maximum motion).
When creating a standing wave on a string that is fixed at both ends, such as in our problem, the boundary conditions necessitate that the ends must be nodes. In the problem solved, a standing wave is formed with two antinodes.
This scenario means the string has one full wavelength because the string length (2 m) matches the wavelength.
Standing waves are integral to understanding how musical instruments produce sound, as they give insight into wave patterns and resulting tones.

Linear Mass Density

Linear mass density, denoted as \( \mu \), is crucial for determining how a wave behaves on a string. It is the mass per unit length of the string.
Calculating linear mass density allows us to determine the speed of the wave, given the string tension.
It is expressed as:

• \( \mu = \frac{m}{L} \)

where \( m \) is the mass of the string, and \( L \) is the length. In the exercise, the string weighing 10g and being 2 meters long resulted in a linear mass density of \( 5 \times 10^{-3} \frac{\text{kg}}{\text{m}} \).
This value plays a pivotal part in understanding the dynamics of waves on strings, directly affecting wave speed.

Tension in String

Tension in the string is a vital concept when examining waves on strings.
It refers to the force exerted along the string, influencing wave speed significantly.
The greater the tension, the faster the wave can travel, leading to changes in frequency and wavelength. Tension is a controllable parameter in musical instruments and other practical applications, making it essential for manipulations of wave properties. In the example provided, a tension of 150 N was applied to the string.
This level of tension, coupled with the string's inherent linear mass density, resulted in a calculated wave speed of 173 m/s.
Understanding and adjusting string tension is crucial for many physical systems where precise wave characteristics are desired.