Question

A string \(2.72 \mathrm{~m}\) long has a mass of \(263 \mathrm{~g}\). The tension in the string is \(36.1 \mathrm{~N}\). What must be the frequency of traveling waves of amplitude \(7.70 \mathrm{~mm}\) in order that the average transmitted power be \(85.5 \mathrm{~W}\) ?

Short answer

The frequency of the wave must be 14.8 Hz.

Step-by-step solution

Calculate the mass per unit length \(\mu\)

Since the mass is given in grams and the length in meters, we need to convert the mass to kg for consistency. Therefore, \(\mu = \frac{263g}{1000g/kg} \times \frac{1}{2.72m} = 0.0967 kg/m\)

Find the wave velocity \(v\)

We can now use the formula \(v = \sqrt{T/\mu}\). Substituting, \(v = \sqrt{36.1N / 0.0967 kg/m} = 19.1 m/s\)

Calculate the frequency \(f\)

Now, rearrange the power formula \(P = 2\pi^2Af^2v^2\) to solve for frequency \(f\): \(f = \sqrt{P / (2\pi^2Av^2)}\). Substitute the given values into this formula: \(f = \sqrt{85.5W / (2\pi^2 * 7.70mm * (19.1 m/s)^2)} = 14.8 Hz\)

Key concepts

Mass per Unit Length

When we talk about waves traveling along a string, the term 'mass per unit length' is crucial for understanding wave dynamics. This term, often represented by the symbol \( \mu \), is defined as the mass of the string divided by its length. Consider it as how much mass is distributed along each meter of the string. In practical terms, a higher mass per unit length would mean a heavier string, which affects how waves will travel along it.

To calculate it, you would take the total mass of the string, ensure it's in kilograms for consistency with units in physics equations, and then divide it by the total length of the string in meters. In our exercise, we converted the mass from grams to kilograms and then divided by the string's length to find a \( \mu \) of 0.0967 kg/m. This value is then used as part of the formula to calculate the wave velocity.

Wave Velocity

Wave velocity, represented by the symbol \( v \), is the speed at which waves travel down the string. The velocity of a wave on a string is determined by two factors: the tension in the string and the mass per unit length. In the world of physics, the formula \( v = \sqrt{T/\mu} \) elegantly relates these factors, indicating how the speed increases with greater tension or decreases with greater mass per length.

In our exercise, after already having calculated \( \mu \) and knowing the tension \( T \), we found the wave velocity to be 19.1 m/s. This gives us the speed of the wave which is essential for further calculations. Understanding this concept is critical because wave velocity affects how quickly information or energy is transmitted along the string.

Frequency of Waves

Frequency, denoted by the symbol \( f \), is the number of waves that pass a given point per second; essentially, it's how often the waves are oscillating. Measured in hertz (Hz), frequency is directly related to the energy a wave can carry. Higher frequency waves on a string will appear closer together and oscillate more quickly.

Using the given formula for power involving amplitude, frequency, and wave velocity, we rearranged the equation to solve for frequency. Our exercise required us to calculate the frequency necessary for a wave to carry a certain amount of power. With the other variables already determined, we found the desired frequency to be 14.8 Hz. Frequency is one of the key characteristics of a wave and alters the wave's energy transmission properties.

Average Transmitted Power

The average transmitted power associated with a traveling wave on a string is a measure of the energy transferred by the wave per unit of time, expressed in watts (W). The power transferred can be crucial when assessing how effectively a wave can perform work, for example, in stringed musical instruments or data transmission cables.

To determine the power in our exercise, we used the formula \( P = 2\pi^2Af^2v^2 \) which incorporates amplitude (\( A \)), frequency (\( f \) ), and wave velocity (\( v \) ). By manipulating this formula, we solved for the required frequency to reach a desired power output. Understanding this gives us insight into the amount of energy the string's vibrations are delivering, which is fundamental for applications that rely on wave energy transfer.