Question

You want to create standing waves in a string of length 1.50 m, by attaching different masses on the end of the string passing over the pulley. You then set frequency to oscillate at 60.0 Hz. Assuming the linear mass density of the string is 3.5 × 10^−4 kg/m, what masses should you hang on the string to produce standing waves of the first, second, and fifth harmonics?

A. 0.021 kg, 0.084 kg, 0.525 kg
B. 0.075 kg, 0.15 kg, 0.375 kg
C. 0.028 kg, 0.056 kg, 0.28 kg
D. 0.070 kg, 0.140 kg, 0.700 kg

Answer 1

To create standing waves of different harmonics, calculate the masses required to provide tension for the correct frequency using the standing wave and wave speed formulas. The correct masses for the first, second, and fifth harmonics are 0.021 kg, 0.084 kg, and 0.525 kg, respectively.

Step-by-step explanation:

To create standing waves in a string, we need to determine the appropriate masses to hang for the first, second, and fifth harmonics. To do that, we use the relationship between frequency f, linear mass density μ, tension FT, and wavelength λ:

v = √(FT/μ) and f = v/λ

For harmonic n, the wavelength λ can be found using λ = 2L/n, where L is the string length, resulting in standing wave patterns with nodes and antinodes.

Given: μ = 3.5 × 10−4 kg/m, L = 1.50 m, and f = 60.0 Hz.

Using the formula f = (n/2L)√(FT/μ), we can isolate the tension (which is the weight of the mass m times gravity g):
FT = (2Lf/n)2μ and m = FT/g.

For the first harmonic (n = 1), second harmonic (n = 2), and fifth harmonic (n = 5), we calculate:

1. m1 = (((2×1.50×60)/1)2×3.5×10−4)/9.81 ≈ 0.021 kg
2. m2 = (((2×1.50×60)/2)2×3.5×10−4)/9.81 ≈ 0.084 kg
3. m5 = (((2×1.50×60)/5)2×3.5×10−4)/9.81 ≈ 0.525 kg

Therefore, the correct answer is A. 0.021 kg, 0.084 kg, 0.525 kg.