Question

If transverse waves on a string travel with a velocity of \(50 \mathrm{~m} / \mathrm{s}\) when the string is under a tension of \(20 \mathrm{~N},\) what tension on the string is required for the waves to travel with a velocity of \(30 \mathrm{~m} / \mathrm{s} ?\) a) \(7.2 \mathrm{~N}\) c) \(33 \mathrm{~N}\) e) \(45 \mathrm{~N}\) b) \(12 \mathrm{~N}\) d) \(40 \mathrm{~N}\) f) \(56 \mathrm{~N}\)

Short answer

Answer: (a) 7.2 N

Step-by-step solution

Write down given information

We are given the following information: Initial wave speed \(v_1 = 50 \mathrm{~m} / \mathrm{s}\) Initial tension \(T_1 = 20 \mathrm{~N}\) Desired wave speed \(v_2 = 30 \mathrm{~m} / \mathrm{s}\) We need to find the desired tension \(T_2\).

Write down wave speed formula

We'll use the wave speed formula: \(v = \sqrt{\frac{T}{\mu}}\), where \(v\) is the wave speed, \(T\) is the tension, and \(\mu\) is the mass per unit length of the string.

Set up the relationship between initial and desired wave speeds and tension

We have the two wave speeds related to two different tensions, \(v_1 = \sqrt{\frac{T_1}{\mu}}\) and \(v_2 = \sqrt{\frac{T_2}{\mu}}\). Divide the first equation by the second equation to eliminate \(\mu\): \(\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}\)

Solve for desired tension

Now we have a relationship we can use to find \(T_2\). First, square both sides of the equation: \(\frac{v_1^2}{v_2^2}=\frac{T_1}{T_2}\) Next, solve for \(T_2\): \(T_2 = T_1 \cdot \frac{v_2^2}{v_1^2}\) Plug the given values for \(v_1\), \(v_2\), and \(T_1\): \(T_2 = 20 \mathrm{N} \cdot \frac{(30 \mathrm{~m} / \mathrm{s})^2}{(50 \mathrm{~m} / \mathrm{s})^2}\) Solve for \(T_2\): \(T_2 = 20 \mathrm{N} \cdot \frac{900}{2500} \Rightarrow T_2 = 20 \mathrm{N} \cdot 0.36 \Rightarrow T_2 = 7.2 \mathrm{~N}\) The required tension for the waves to travel with a velocity of \(30 \mathrm{~m} / \mathrm{s}\) is \(7.2 \mathrm{~N}\), which corresponds to answer (a).

Key concepts

Wave Speed

Wave speed is simply the distance a wave travels per unit of time, or how fast the wave moves through a medium. In physics, wave speed is an essential concept because it ties together the physical properties of the medium and the wave properties themselves, such as frequency and wavelength.
To calculate wave speed, we use the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( v \) is the wave speed, \( T \) is the tension in the medium (such as a string), and \( \mu \) is the mass per unit length.
For example, suppose a wave travels at 50 m/s on a string with a tension of 20 N. If the tension changes, the wave speed will also change, reflecting how tension in strings affects wave propagation.

Tension in Strings

Tension in strings significantly impacts wave behavior, particularly how fast the waves travel along the string. The tension refers to the pulling force applied along the string, which influences the string's tightness.
When a string is pulled tighter (increased tension), waves move faster across it. Conversely, if tension decreases, the wave speed lowers. This relationship stems from the inherent characteristic of mechanical waves: they require a medium to travel through, and that medium's properties, like tension, affect the wave's speed.

• Positive correlation: Higher tension, higher wave speed.
• Negative correlation: Lower tension, slower wave speed.

This concept is crucial in various applications, such as musical instruments, where string tension dictates the pitch of sound produced.

Transverse Waves

Transverse waves are waves where the direction of oscillation is perpendicular to the direction of the wave's movement. A common example includes waves on a string or the ripples on the water surface.
These waves can be visualized as a series of crests and troughs moving through the medium. For a transverse wave on a string, the motion of the string is up and down, while the wave propagates horizontally.
The wave speed in transverse waves, especially on strings, can be influenced by:

• The tension in the string.
• The mass per unit length of the string.

The higher the speed of transverse waves, the more energy is transferred in a given amount of time. Understanding transverse waves helps unveil how energy is distributed in waves, which is pivotal in fields such as acoustics and communication technologies.

Mass per Unit Length

Mass per unit length, denoted as \( \mu \), refers to the amount of mass contained in a segment of the string. It is calculated by dividing the total mass of the string by its length.
This parameter is crucial in determining the wave speed in strings because it directly influences how quickly waves can propagate. The formula showing this relationship is \( v = \sqrt{\frac{T}{\mu}} \), indicating that for a fixed tension, a string with lower mass per unit length will allow faster wave speeds.

• Influences wave properties: Less mass, faster wave speed for a given tension.
• Determines wave performance in musical instruments, engineering applications, etc.

Understanding mass per unit length helps optimize applications where wave speed regulation is vital, ensuring the desired performance and efficiency in real-world applications.