Question

In an acoustics experiment, a piano string with a mass of \(5.00 \mathrm{~g}\) and a length of \(70.0 \mathrm{~cm}\) is held under tension by running the string over a frictionless pulley and hanging a \(250 .-\mathrm{kg}\) weight from it. The whole system is placed in an elevator. a) What is the fundamental frequency of oscillation for the string when the elevator is at rest? b) With what acceleration and in what direction (up or down) should the elevator move for the string to produce the proper frequency of \(440 . \mathrm{Hz}\) corresponding to middle A?

Short answer

(b) If the elevator accelerates, what must its acceleration be to produce a fundamental frequency of \(440 \text{ Hz}\) (middle A)? Solution: Step 1: The tension in the string due to the weight (at rest) is calculated as: \(T= m_{w}g = 250 .-\mathrm{kg} \times 9.81 \mathrm{m/s^2} = 2452.5 \mathrm{N}\) Step 2: The mass per unit length of the string (\(\mu\)) is calculated as: \(\mu = \frac{m_s}{L} = \frac{0.05 \mathrm{kg}}{1.8 \mathrm{m}} = 0.0278 \mathrm{kg/m}\) Now, we calculate the fundamental frequency (\(f_1\)) when the elevator is at rest: \(f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}} = \frac{1}{2(1.8)}\sqrt{\frac{2452.5}{0.0278}} = 217.47 \mathrm{Hz}\) The fundamental frequency when the elevator is at rest is \(217.47 \mathrm{Hz}\). Step 3: To find the new tension needed for the frequency of 440 Hz, we calculate: \(T_{new} = (\frac{2Lf_2}{\sqrt{\mu}})^2 = (\frac{2(1.8)(440)}{\sqrt{0.0278}})^2 = 9809.63 \mathrm{N}\) Step 4: Finally, we calculate the acceleration of the elevator: \(a = \frac{T_{new} - T}{m_{w} + m_s} = \frac{9809.63 - 2452.5}{250 + 0.05} = 2.934 \mathrm{m/s^2}\) The acceleration of the elevator must be approximately \(2.934 \mathrm{m/s^2}\) for the piano string to produce a fundamental frequency of \(440 \mathrm{Hz}\) (middle A).

Step-by-step solution

Calculate the tension in the string due to the weight

To calculate the tension in the string when the elevator is at rest, we can use the weight of the \(250 .-\mathrm{kg}\) mass hanging from it. The tension will be equal to the gravitational force acting on the hanging mass: \(T = m_{w}g\) where \(m_{w}\) is the mass of the weight (\(250 .-\mathrm{kg}\)) and \(g\) is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)).

Find the fundamental frequency when the elevator is at rest

To find the fundamental frequency when the elevator is at rest, we need to use the tension, mass, and length of the piano string. The formula for the fundamental frequency is: \(f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\) where \(f_1\) is the fundamental frequency, \(L\) is the length of the piano string, \(T\) is the tension in the string, and \(\mu\) is the mass per unit length of the string. To find \(\mu\), simply divide the mass of the string by its length: \(\mu = \frac{m_s}{L}\) where \(m_s\) is the mass of the piano string and \(L\) is its length. Plug in the values and solve for \(f_1\).

Calculate the new tension needed for the frequency of 440 Hz

Now, we need to find the tension needed for the string to produce the proper frequency of 440 Hz. We will use the formula for the fundamental frequency: \(f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\) Since we are solving for the new tension \(T_{new}\), we can rearrange the formula: \(T_{new} = (\frac{2Lf_2}{\sqrt{\mu}})^2\) Plug in the values and solve for \(T_{new}\).

Calculate the acceleration of the elevator

To find the acceleration of the elevator (\(a\)), we need to look at the difference in tension between the tension when the elevator is at rest and the new tension needed for the frequency of 440 Hz. We can use Newton's second law of motion to relate the tension and acceleration: \(T_{new} - T = (m_{w} + \frac{m_s}{L}L)a\) Solve for \(a\) using the previously calculated values of \(T_{new}\), \(T\), \(m_w\), and \(m_s\). Positive \(a\) values represent that the elevator moves upwards, and negative \(a\) values represent downward motion of the elevator.