Question

Tension is maintained in a string by attaching one end to a wall and by hanging a \(2.20-\mathrm{kg}\) object from the other end of the string after it passes over a pulley that is \(2.00 \mathrm{m}\) from the wall. The string has a mass per unit length of \(3.55 \mathrm{mg} / \mathrm{m} .\) What is the fundamental frequency of this string?

Short answer

Answer: The fundamental frequency of the string is approximately 26.38 Hz.

Step-by-step solution

Calculate the tension in the string

Tension in the string is equal to the weight of the hanging object, which can be calculated as \(T = mg\), where \(T\) is the tension, \(m\) is the mass, and \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{m/s^2}\)). Our mass is given as \(2.20 \text{ kg}\), so we have \(T = (2.20 \text{ kg})(9.81 \mathrm{ m/s^2})\). Solving for T, we get \(T = 21.582 \text{ N}\).

Determine the mass per unit length of the string

We are given the mass per unit length of the string as \(3.55 \mathrm{mg / m}\). We need to convert this to \(\mathrm{kg / m}\) for calculations. \(3.55 \mathrm{mg} = 3.55 \times 10^{-3} \mathrm{g}\), and since \(1 \mathrm{g} = 10^{-3} \mathrm{kg}\), we have \(3.55 \times 10^{-3} \mathrm{g} = 3.55 \times 10^{-6} \mathrm{kg}\). So, the mass per unit length of the string is \(3.55 \times 10^{-6} \mathrm{kg / m}\).

Use the fundamental frequency formula for a string

The fundamental frequency for a string is given by the formula \(f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\), where \(f\) is the fundamental frequency, \(L\) is the length of the string, \(T\) is the tension, and \(\mu\) is the mass per unit length. In this case, the distance of the pulley from the wall, \(2.00 \text{ m}\), is equal to the length of the string, so \(L = 2.00 \text{ m}\). Using the values of tension and mass per unit length that we found previously, the fundamental frequency can be calculated as \(f = \frac{1}{2(2.00 \text{ m})} \sqrt{\frac{21.582 \text{ N}}{3.55 \times 10^{-6} \mathrm{kg / m}}}\).

Calculate the fundamental frequency

Now we can calculate the fundamental frequency by solving the equation: \(f = \frac{1}{4.00 \text{ m}} \sqrt{\frac{21.582 \text{ N}}{3.55 \times 10^{-6} \mathrm{kg / m}}}\) \(f = 26.38 \text{ Hz}\) (rounded to 2 decimal places) So, the fundamental frequency of this string is approximately \(26.38 \text{ Hz}\).