Two waves traveling in opposite directions along a string fixed at both ends create a standing wave described by \(y(x, t)=1.00 \cdot 10^{-2} \sin (25 x) \cos (1200 t) .\) The string has a linear mass density of \(0.01 \mathrm{~kg} / \mathrm{m},\) and the tension in the string is supplied by a mass hanging from one end. If the string vibrates in its third harmonic, calculate (a) the length of the string, (b) the velocity of the waves, and (c) the mass of the hanging mass.

Short Answer

Expert verified
Answer: The length of the string is \(L = 0.377\,\mathrm{m}\), the velocity of the waves is \(v = 48\,\mathrm{m/s}\), and the mass of the hanging mass is \(m = 11.83\,\mathrm{kg}\).

Step by step solution

01

Calculate the Angular Wave Number (k)

For a given standing wave, the angular wave number (k) can be calculated from the equation: \(y(x, t) = A\sin(kx)\cos(\omega t)\) Comparing this to the given equation: \(y(x, t) = 1.00 \times 10^{-2} \sin (25 x) \cos (1200 t)\) We have \(k = 25 \frac{rad}{m}\).
02

Calculate the length of the string (L)

In a standing wave vibrating in its nth harmonic, the length of the string (L) can be found using the following formula: \(L = \frac{n\pi}{k}\) Substituting the values for n and k: \(L = \frac{3\pi}{25} = 0.377 \,\mathrm{m}\)
03

Calculate the Angular Frequency (\(\omega\))

From the given standing wave equation, we can obtain the angular frequency \(\omega\): \(y(x, t) = 1.00 \times 10^{-2} \sin (25 x) \cos (1200 t)\) \(\omega = 1200\,\mathrm{rad/s}\)
04

Calculate the velocity of the waves (v)

The relationship between the angular wave number (k) and the angular frequency (\(\omega\)) is given by: \(v = \frac{\omega}{k}\) Plugging in the values for \(\omega\) and k, we get: \(v = \frac{1200\,\mathrm{rad/s}}{25\,\mathrm{rad/m}} = 48\,\mathrm{m/s}\)
05

Calculate the mass of the hanging mass (m)

The tension (T) in the string is the force exerted by the hanging mass (m), which is given by: \(T = mg\) The wave speed v can also be calculated using the formula: \(v = \sqrt{\frac{T}{\mu}}\) Solving for m, we get: \(m = \frac{\mu v^2}{g}\) Substituting the known values: \(m = \frac{0.01\,\mathrm{kg/m} \times (48\,\mathrm{m/s})^2}{9.81\,\mathrm{m/s^2}} = 11.83\,\mathrm{kg}\) Therefore, the solutions to the problem are: (a) The length of the string is \(L = 0.377\,\mathrm{m}\). (b) The velocity of the waves is \(v = 48\,\mathrm{m/s}\). (c) The mass of the hanging mass is \(m = 11.83\,\mathrm{kg}\).

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