Question

Two waves traveling in opposite directions along a string fixed at both ends create a standing wave described by \(y(x, t)=\) \(1.00 \cdot 10^{-2} \sin (25.0 x) \cos (1200 . t) .\) The string has a linear mass density of \(0.0100 \mathrm{~kg} / \mathrm{m},\) 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

In the given situation, the length of the string (L) is $\frac{3\pi}{25.0}$ m, the velocity of the waves (v) is 48 m/s, and the mass of the hanging object (m) is 2.35 kg.

Step-by-step solution

a) Finding the length of the string

First, let's analyze the equation for the standing wave: $y(x, t) = 1.00 \cdot 10^{-2} \sin (25.0 x) \cos (1200 . t)$. We see that the term responsible for the spatial distribution of the wave is \(\sin(25.0x)\). Since the string is in its third harmonic, this means that it has \(3\) nodes including the fixed ends. Thus, the length of the string \(L\) corresponds to \(3/2\) wavelengths. We can determine the wavelength \(\lambda\) by finding the argument of the sine function: \(25.0x = kx = \frac{2\pi}{\lambda} x\), where \(k\) is the wave number. Thus, \(\frac{2\pi}{\lambda} = 25.0\). Solving for \(\lambda\), we get: \(\lambda = \frac{2\pi}{25.0}\). Now we can use this to find the length of the string \(L\): \(L = \frac{3}{2} \lambda = \frac{3}{2} \cdot \frac{2\pi}{25.0} = \frac{3\pi}{25.0}\).

b) Finding the velocity of the waves

The term responsible for the time dependence is \(\cos(1200.t)\). The coefficient of \(t\) is the angular frequency \(\omega\), so \(\omega = 1200\) radians/s. To find the linear frequency \(f\), we can use the equation \(\omega = 2\pi f\). Solving for \(f\), we get: \(f = \frac{\omega}{2\pi} = \frac{1200}{2\pi}\) Hz. Now that we have the frequency \(f\) and the wavelength \(\lambda\), we can find the velocity \(v\) of the waves using the equation \(v = f\lambda\): \(v = f \cdot \lambda = \frac{1200}{2\pi} \cdot \frac{2\pi}{25.0} = 48\) m/s.

c) Finding the mass of the hanging mass

To find the tension in the string \(T\), we can use the formula \(T = \mu v^2\), where \(\mu\) is the linear mass density: \(T = \mu v^2 = (0.0100 \,\mathrm{kg/m}) (48 \,\mathrm{m/s})^2 = 23.04\) N. The tension \(T\) is equal to the weight of the hanging mass \(mg\), where \(m\) is the mass and \(g\) is the gravitational acceleration. We can solve for the mass \(m\): \(m = \frac{T}{g} = \frac{23.04\,\mathrm{N}}{9.81\,\mathrm{m/s^2}} = 2.35\) kg. So, the mass of the hanging mass is 2.35 kg.