Velocity of Transverse Waves: Videos & Practice Problems

Waves on strings can be analyzed using the wave speed equation, v=λf, and a specific equation for wave speed, v = Tμ, where T is tension and μ is mass density. Changes in tension or mass directly affect wave speed, while frequency changes impact wavelength without altering speed. Understanding these relationships is crucial for solving wave problems effectively.

Velocity of Waves on a String

Waves can be classified into two main types: transverse and longitudinal. The wave speed equation, given by v = \lambda f, applies to all types of waves, where v represents wave speed, \lambda is the wavelength, and f is the frequency. However, when dealing with transverse waves on strings, additional considerations regarding the physical properties of the string are necessary to accurately determine wave speed.

For transverse waves on strings, the wave speed can be calculated using the formula:

v = \sqrt{\frac{T}{\mu}}

In this equation, T represents the tension in the string, and \mu is the mass per unit length of the string, also known as the mass density. The mass density can be calculated using the formula:

\mu = \frac{m}{l}

where m is the mass of the string and l is its length.

To illustrate this, consider a string with a tension of 100 Newtons, a mass of 0.5 kilograms, and a length of 1.2 meters. The wavelength is given as 15 centimeters (0.15 meters). First, we calculate the mass density:

\mu = \frac{0.5 \, \text{kg}}{1.2 \, \text{m}} = 0.4167 \, \text{kg/m}

Next, we substitute the values into the wave speed equation:

v = \sqrt{\frac{100 \, \text{N}}{0.4167 \, \text{kg/m}}} \approx 15.4 \, \text{m/s}

Now that we have the wave speed, we can find the frequency using the original wave speed equation:

f = \frac{v}{\lambda} = \frac{15.4 \, \text{m/s}}{0.15 \, \text{m}} \approx 103 \, \text{Hz}

This example demonstrates the importance of understanding both the general wave speed equation and the specific considerations for waves on strings, allowing for accurate calculations of frequency and wave speed based on the physical properties of the medium.

A rope with length 2.50 m and mass 0.10 kg is stretched and pulled to create transverse waves of frequency 40.0 Hz and wavelength of 0.750 m. How much tension is exerted on the rope?

• A. F_T = 36 N
• B. F_T = 6 N
• C. F_T = 1.2 N
• D. F_T = 22,500 N

Answer: (A)

Example 1

In this scenario, we are exploring the concept of wave propagation along a rope. When you create a wave pulse by whipping the rope up and down, the wave travels towards your friend at a certain speed, denoted as \( v \). To determine the time it takes for the wave pulse to reach your friend, we need to calculate the time interval, represented as \( \Delta t \).

The relationship between wave speed, displacement, and time is given by the equation:

\( v = \frac{\Delta x}{\Delta t} \)

Here, \( \Delta x \) represents the length of the rope, which in this case is 7.5 meters. Rearranging the equation allows us to express the time interval as:

\( \Delta t = \frac{\Delta x}{v} \)

Next, we need to determine the wave speed \( v \). For waves on strings, the speed can be calculated using the formula:

\( v = \sqrt{\frac{T}{\mu}} \)

where \( T \) is the tension in the rope and \( \mu \) is the mass per unit length of the rope. In this example, the tension \( T \) is 30 newtons, and the mass of the rope is 0.5 kg with a length of 7.5 meters. Thus, the mass per unit length \( \mu \) can be calculated as:

\( \mu = \frac{0.5 \, \text{kg}}{7.5 \, \text{m}} = 0.0667 \, \text{kg/m} \)

Substituting the values into the wave speed equation gives:

\( v = \sqrt{\frac{30 \, \text{N}}{0.0667 \, \text{kg/m}}} \approx 21.2 \, \text{m/s} \)

Now that we have the wave speed, we can find the time it takes for the wave pulse to travel the length of the rope:

\( \Delta t = \frac{7.5 \, \text{m}}{21.2 \, \text{m/s}} \approx 0.35 \, \text{s} \)

Therefore, it takes approximately 0.35 seconds for the wave pulse to travel from you to your friend along the rope.

Solving Problems with Waves on Strings

In the study of wave mechanics, understanding how various factors influence wave speed, frequency, and wavelength is crucial. The wave speed on a string can be determined using the equation:

\( v = \sqrt{\frac{F_T}{\mu}} \)

where \( v \) is the wave speed, \( F_T \) is the tension in the string, and \( \mu \) is the mass per unit length (mass density) of the string. This relationship indicates that changes in tension or mass density will directly affect the wave speed.

For instance, if the tension in the string is quadrupled, the new tension \( F_T' \) can be calculated as:

\( F_T' = 4 \times F_T \)

Using the updated tension in the wave speed equation, the new wave speed \( v' \) can be calculated. If the initial tension was 98 N, the new tension becomes 392 N, leading to a new wave speed of:

\( v' = \sqrt{\frac{392}{\mu}} \)

Assuming the mass density remains constant at 2 kg/m, the new wave speed is found to be 14 m/s, which is double the initial speed of 7 m/s. This demonstrates that quadrupling the tension results in a doubling of the wave speed.

Next, consider the effect of changing the frequency of the oscillator. If the frequency is doubled, the new frequency \( f' \) becomes:

\( f' = 2 \times f \)

In this case, the frequency increases from 35 Hz to 70 Hz. However, it is important to note that the wave speed \( v \) remains unchanged because it is determined by the physical properties of the medium (tension, mass density, and length), not the oscillator frequency. Therefore, the wave speed remains at 7 m/s.

To find the new wavelength \( \lambda' \), we can use the relationship between wave speed, frequency, and wavelength:

\( v = \lambda f \)

Rearranging this gives:

\( \lambda = \frac{v}{f} \)

Substituting the values, we find:

\( \lambda' = \frac{7 \, \text{m/s}}{70 \, \text{Hz}} = 0.1 \, \text{m} \)

This indicates that as the frequency increases, the wavelength decreases, resulting in more closely spaced waves. Initially, the wavelength was 0.2 m, and after doubling the frequency, it is now 0.1 m.

In summary, changes in tension affect wave speed directly, while changes in frequency affect wavelength without altering wave speed. Understanding these relationships is essential for solving problems related to wave motion on strings.

An oscillating blade creates waves on a string. If the amplitude of the wave doubles, what happens to the wavelength λ and v?

• A. λ and v remain the same
• B. λ doubles, v remains the same
• C. λ remains the same, v doubles

Answer: (A)

Example 2

In this problem, we explore the relationship between wave speed, tension, and mass per unit length in a string. The initial wave speed, denoted as \( v \), is given as 20 meters per second. The wave speed in a string can be calculated using the formula:

\[ v = \sqrt{\frac{F_t}{\mu}} \]

where \( F_t \) represents the tension in the string and \( \mu \) is the mass per unit length of the string. When the tension is cut in half, we need to determine the new wave speed, denoted as \( v_{\text{new}} \).

Since the mass per unit length remains constant, we can express the new wave speed as:

\[ v_{\text{new}} = \sqrt{\frac{F_{t \, \text{new}}}{\mu}} \]

Given that the new tension \( F_{t \, \text{new}} \) is half of the original tension, we can substitute this into the equation:

\[ v_{\text{new}} = \sqrt{\frac{\frac{1}{2} F_t}{\mu}} \]

By extracting the square root of \( \frac{1}{2} \), we can rewrite the equation as:

\[ v_{\text{new}} = \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{F_t}{\mu}} \]

Since we know that \( \sqrt{\frac{F_t}{\mu}} = 20 \) m/s, we can substitute this value into the equation:

\[ v_{\text{new}} = \sqrt{\frac{1}{2}} \cdot 20 \]

Calculating this gives:

\[ v_{\text{new}} = 20 \cdot \frac{1}{\sqrt{2}} \approx 14.1 \, \text{m/s} \]

This result indicates that the new wave speed decreases to approximately 14.1 meters per second when the tension is halved. This outcome aligns with our understanding that reducing the tension in a string results in a lower wave speed, though not in a linear fashion due to the square root relationship in the wave speed formula.

Here’s what students ask on this topic:

What is the equation for the velocity of transverse waves on a string?

The velocity of transverse waves on a string is given by the equation:

$v=\sqrt{\frac{T}{\mu }}$

where $v$ is the wave speed, $T$ is the tension in the string, and $\mu$ is the mass per unit length (mass density) of the string.

How does tension affect the speed of a transverse wave on a string?

The speed of a transverse wave on a string is directly related to the square root of the tension in the string. According to the equation:

$v=\sqrt{\frac{T}{\mu }}$

if the tension $T$ increases, the wave speed $v$ also increases. Specifically, if the tension is quadrupled, the wave speed will double.

What is the relationship between wave speed, frequency, and wavelength for transverse waves?

The relationship between wave speed $v$, frequency $f$, and wavelength $\lambda$ for transverse waves is given by the equation:

$v=\lambda f$

This means that the wave speed is the product of the wavelength and the frequency. If either the frequency or the wavelength changes, the other must adjust to keep the wave speed constant, assuming the medium does not change.

How do you calculate the mass density of a string?

The mass density $\mu$ of a string is calculated by dividing the mass $m$ of the string by its length $l$:

$\mu =\frac{m}{l}$

This value is essential for determining the wave speed on the string using the wave speed equation.

What happens to the wavelength if the frequency of a transverse wave is doubled?

If the frequency $f$ of a transverse wave is doubled, the wavelength $\lambda$ is halved, assuming the wave speed $v$ remains constant. This relationship is described by the equation:

$v=\lambda f$

Since the wave speed is the product of wavelength and frequency, if the frequency increases, the wavelength must decrease proportionally to maintain the same wave speed.